LeetCode: Tags-[Array], Difficulty-[Medium]
Tags: Array
Difficulty: Medium
11. Container With Most Water: https://leetcode.com/problems/container-with-most-water/
n条纵线,找出两条线使得容器最大:(题意:假设ai,aj Cmax =min(ai,aj)*(j-i));
暴力解法: (内外for循环; 外循环遍历, 内循环遍历i后的数, 嵌套求最值 max = Math.max(max,min([i],[j]))*(j-i) ps:LeetCode最后大数据TLE,LintCode可以通过)
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1 public class Solution { 2 public int maxArea(int[] height) { 3 int max = 0; 4 for (int i = 0; i < height.length; i++) { 5 for (int j = i; j < height.length; j++) { 6 max = Math.max(max,Math.min(height[j],height[i])*(j - i)); 7 } 8 } 9 return max; 10 } 11 }
解法2:<Two Pointers> (首尾双指针遍历 left,right, 记录max, 保留[left]/[right]中大的一个,另一边指针移动;)
(0.int area; 1.left,right,max; 2.while(<=)max=max(,area); 3.if([left]<)left++); (Q: 可以嵌套for循环的就可以双指针??Todo;)
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1 public class Solution { 2 /** 3 * @param heights: an array of integers 4 * @return: an integer 5 */ 6 int computeArea(int left, int right, int[] heights) { 7 return (right-left)*Math.min(heights[left], heights[right]); 8 } 9 10 public int maxArea(int[] heights) { 11 // write your code here 12 int left = 0, ans= 0 ; 13 int right = heights.length - 1; 14 while(left <= right) { 15 ans = Math.max(ans,computeArea(left, right, heights)); 16 if(heights[left]<=heights[right]) 17 left++; 18 else 19 right--; 20 } 21 return ans; 22 } 23 }
15. 3Sum: https://leetcode.com/problems/3sum/
找出数组中所有3数之和为零的数:
解法:(遍历,第一个数的相反数作为后两个数的和,二分遍历后两个数 1.排序 2.for(0,n-2)遍历 3.if(i=0 || [i]=[i-1]) 跳过重复元素 4.lo,high,sum 5.while(<)if等sum则add,while(<&&=)lo++,while(<&&=)hi--跳过重复元素; lo++ hi-- 搜索下一对2SUM; 6.if(<sum)那么lo++,else()hi--; )
(1.sort,rst; 2.for(n-2)if(0||!=); 3.lo=i+1,hi,sum; 4.while() if(=)tmp,add,while(<&=)++while(>&=)--; 5.if(<)++(>)-- 6.return;)
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1 public class Solution { 2 public List<List<Integer>> threeSum(int[] nums) { 3 Arrays.sort(nums); 4 List<List<Integer>> rst = new ArrayList<List<Integer>>(); 5 for (int i = 0; i < nums.length - 2; i++) { 6 if (i == 0 || (i > 0 && nums[i] != nums[i - 1])){ 7 int lo = i + 1, hi = nums.length - 1, sum = 0 - nums[i]; 8 while (lo < hi) { 9 if (nums[lo] + nums[hi] == sum) { 10 List<Integer> tmp = new ArrayList<Integer>(); 11 tmp.add(nums[i]); 12 tmp.add(nums[lo]); 13 tmp.add(nums[hi]); 14 rst.add(tmp); 15 while (lo < hi && nums[lo] == nums[lo + 1]) lo++; 16 while (lo < hi && nums[hi] == nums[hi - 1]) hi--; 17 lo++; hi--; 18 } else if (nums[lo] + nums[hi] < sum) lo++; 19 else hi--; 20 } 21 } 22 } 23 return rst; 24 } 25 }
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1 public List<List<Integer>> threeSum(int[] num) { 2 Arrays.sort(num); 3 List<List<Integer>> res = new LinkedList<>(); 4 for (int i = 0; i < num.length-2; i++) { 5 if (i == 0 || (i > 0 && num[i] != num[i-1])) { 6 int lo = i+1, hi = num.length-1, sum = 0 - num[i]; 7 while (lo < hi) { 8 if (num[lo] + num[hi] == sum) { 9 res.add(Arrays.asList(num[i], num[lo], num[hi])); 10 while (lo < hi && num[lo] == num[lo+1]) lo++; 11 while (lo < hi && num[hi] == num[hi-1]) hi--; 12 lo++; hi--; 13 } else if (num[lo] + num[hi] < sum) lo++; 14 else hi--; 15 } 16 } 17 } 18 return res; 19 }
16.3Sum Closest: https://leetcode.com/problems/3sum-closest/
找出数组中三数之和最接近target,返回三数之和:
解法:(遍历第一个数,后两个数二分遍历,比较rst和遍历出的sum的与tar的差值大小;根据sum和tar的差值决定lo和hi的变化)(和我自己思路类似,但是我卡在了lo和hi的变化上,不能用lo++hi——,这样会导致有的组合被跳过 [a,b,c,d]->[a]+[d]->[b]+[c] 漏掉了ab,ac,bd,cd 所以应该根据sum和tar的大小关系确定lo和hi的变化;)(用到方法:Math.abs()取绝对值)
(1.if()-1; 2.sort,rst; 3.for(n-2) if()lo,hi; 4.while()sum if(Math.abs<)rst=sum; 5.if()lo++;if()hi-- 6.return;) (ps:while里的语句跳过重复,也可以不用)
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1 public class Solution { 2 public int threeSumClosest(int[] nums, int target) { 3 if (nums == null || nums.length < 3) { 4 return -1; 5 } 6 Arrays.sort(nums); 7 int rst = nums[0] + nums[1] + nums[2]; 8 for (int i = 0; i < nums.length - 2; i++) { 9 if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) { 10 int lo = i + 1, hi = nums.length - 1; 11 while (lo < hi){ 12 int sum = nums[lo] + nums[hi] + nums[i]; 13 if (Math.abs(rst - target) > Math.abs(sum - target)) 14 rst = sum; 15 if (sum < target) 16 lo++; 17 else hi--; 18 } 19 } 20 } 21 return rst; 22 } 23 }
18. 4Sum:https://leetcode.com/problems/4sum/
我的解法:(与3SUM解法相同,两趟for循环遍历第一个值和第二个值,注意每趟遍历依然要去重!且每趟for循环的首位要保留;[00000] i=0,j=1去重时不能把j=1也去了)
(1.rst,sort; 2.for(n-3)if(||&&); for(i+!,n-2)if(||&&); 3.lo,hi while(); 4.if(=)tmp,add,while(&); 5.else if(<) else(>) 6.return;)
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1 public class Solution { 2 public List<List<Integer>> fourSum(int[] nums, int target) { 3 List<List<Integer>> rst = new ArrayList<List<Integer>>(); 4 Arrays.sort(nums); 5 for (int i = 0; i < nums.length - 3; i++) { 6 if (i == 0 || (i !=0 && nums[i] != nums[i - 1])){ 7 for (int j = i + 1; j < nums.length - 2; j++) { 8 if (j == i + 1 || (j != i + 1 && nums[j] != nums[j - 1])){ 9 int lo = j + 1, hi = nums.length - 1; 10 while (lo < hi) { 11 if (nums[lo] + nums[hi] + nums[i] + nums[j] == target) { 12 List<Integer> tmp = new ArrayList<Integer>(); 13 tmp.add(nums[i]); tmp.add(nums[j]); 14 tmp.add(nums[lo]); tmp.add(nums[hi]); 15 rst.add(tmp); 16 while (lo < hi && nums[lo] == nums[lo + 1]) lo++; 17 while (lo < hi && nums[hi] == nums[hi - 1]) hi--; 18 lo++; hi--; 19 } else if (nums[i] + nums[j] + nums[lo] + nums[hi] < target) 20 lo++; 21 else hi --; 22 } 23 } 24 } 25 } 26 } 27 return rst; 28 } 29 }
if也可以有其他写法:(continue)
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1 public class Solution { 2 public List<List<Integer>> fourSum(int[] num, int target) { 3 ArrayList<List<Integer>> ans = new ArrayList<>(); 4 if(num.length<4)return ans; 5 Arrays.sort(num); 6 for(int i=0; i<num.length-3; i++){ 7 if(i>0&&num[i]==num[i-1])continue; 8 for(int j=i+1; j<num.length-2; j++){ 9 if(j>i+1&&num[j]==num[j-1])continue; 10 int low=j+1, high=num.length-1; 11 while(low<high){ 12 int sum=num[i]+num[j]+num[low]+num[high]; 13 if(sum==target){ 14 ans.add(Arrays.asList(num[i], num[j], num[low], num[high])); 15 while(low<high&&num[low]==num[low+1])low++; 16 while(low<high&&num[high]==num[high-1])high--; 17 low++; 18 high--; 19 } 20 else if(sum<target)low++; 21 else high--; 22 } 23 } 24 } 25 return ans; 26 } 27 }
31. Next Permutation: https://leetcode.com/problems/next-permutation/
找出下一个排列:[1,2,3,1]->[1,3,1,2]; [1,3,5,4,3,2,1]->[1,4,1,2,3,3,5]
解法:(1.从后往前遍历,找到第一个不满足降序的i; 2.从后往前遍历 找出i后第一个比i大的数并swap; 3.将i后的数(已经是倒序)reverse反转; 这样能保证变化范围最小)
(我的代码,有问题。。暂时没找到原因……)
错!误!代!码!:原因swap(i,j)是按值传递,swap(nums[i],nums[j])后对数组没有变化;swap(int[] nums, int i, int j)是引用传递,将数组作为方法的参数,可以改变数组
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1 public class Solution { 2 public void nextPermutation(int[] nums) { 3 if (nums == null || nums.length == 0) { 4 return; 5 } 6 int i = nums.length - 1; 7 for (; i >= 1; i--) { 8 if (nums[i] > nums[i - 1]) { 9 break; 10 } 11 } 12 if (i != 0) { 13 int j = nums.length - 1; 14 for (; j >= i; j--) { 15 if (nums[j] > nums[i - 1]) { 16 swap(nums[i - 1],nums[j]); 17 break; 18 } 19 } 20 } 21 reverse(nums,i); 22 } 23 void swap(int i,int j) { 24 int tmp = i; 25 i = j; 26 j = tmp; 27 } 28 void reverse(int[] nums, int start) { 29 int end = nums.length - 1; 30 for (; start < end; start++,end--) { 31 swap(nums[start],nums[end]); 32 } 33 } 34 }
正确代码:(利用break记录指针最后的位置)
(1.if; 2.i for(0)if()break; 3.if(!=)j,for(>=) if(i-1)swap break; 4.reverse(i); 5.swap([]),reverse([],start))
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1 public class Solution { 2 public void nextPermutation(int[] nums) { 3 if (nums == null || nums.length == 0) { 4 return; 5 } 6 int i = nums.length - 1; 7 for (; i > 0; i--) { 8 if (nums[i] > nums[i - 1]) { 9 break; 10 } 11 } 12 if (i != 0) { 13 int j = nums.length - 1; 14 for (; j >= i; j--) { 15 if (nums[j] > nums[i - 1]) { 16 swap(nums, i - 1, j); 17 break; 18 } 19 } 20 } 21 reverse(nums, i); 22 } 23 void swap(int[] nums, int i, int j) { 24 int tmp = nums[i]; 25 nums[i] = nums[j]; 26 nums[j] = tmp; 27 } 28 void reverse(int[] nums, int start) { 29 int end = nums.length - 1; 30 for (; start < end; start++,end--) { 31 swap(nums,start,end); 32 } 33 } 34 }
类似的:
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1 public class Solution { 2 public void nextPermutation(int[] nums) { 3 // find two adjacent elements, n[i-1] < n[i] 4 int i = nums.length - 1; 5 for (; i > 0; i --) { 6 if (nums[i] > nums[i-1]) { 7 break; 8 } 9 } 10 if (i != 0) { 11 // swap (i-1, x), where x is index of the smallest number in [i, n) 12 int x = nums.length - 1; 13 for (; x >= i; x --) { 14 if (nums[x] > nums[i-1]) { 15 break; 16 } 17 } 18 swap(nums, i - 1, x); 19 } 20 reverse(nums, i, nums.length - 1); 21 } 22 23 void swap(int[] a, int i, int j) { 24 int t = a[i]; 25 a[i] = a[j]; 26 a[j] = t; 27 } 28 // reverse a[i, j] 29 void reverse(int[] a, int i, int j) { 30 for (; i < j; i ++, j --) { 31 swap(a, i, j); 32 } 33 } 34 }
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1 public class Solution { 2 public void nextPermutation(int[] nums) { 3 if(nums.length<=1){ 4 return; 5 } 6 7 int i= nums.length-1; 8 9 for(;i>=1;i--){ 10 11 if(nums[i]>nums[i-1]){ //find first number which is smaller than it's after number 12 break; 13 } 14 } 15 16 if(i!=0){ 17 swap(nums,i-1); //if the number exist,which means that the nums not like{5,4,3,2,1} 18 } 19 20 reverse(nums,i); 21 } 22 23 private void swap(int[] a,int i){ 24 for(int j = a.length-1;j>i;j--){ 25 if(a[j]>a[i]){ 26 int t = a[j]; 27 a[j] = a[i]; 28 a[i] = t; 29 break; 30 } 31 } 32 } 33 34 private void reverse(int[] a,int i){//reverse the number after the number we have found 35 int first = i; 36 int last = a.length-1; 37 while(first<last){ 38 int t = a[first]; 39 a[first] = a[last]; 40 a[last] = t; 41 first ++; 42 last --; 43 } 44 } 45 46 }
34. Search for a Range: https://leetcode.com/problems/search-for-a-range/
找出有序数组中tar的首尾位置:
九章解法:(两次二分查找,注意while循环条件是start+1<end; 一般涉及范围问题、有重复值问题都应该用这个条件;对应的,下面的mid=start,end;并且要在最后判断边界值start和end。这样可以模板化,避免判断最终边界值start、mid、end混乱、错误的问题)
(start,end,while(+1<)mid,mid if([]=)else if([]=))
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1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 if (A.length == 0) { 4 return new int[]{-1, -1}; 5 } 6 7 int start, end, mid; 8 int[] bound = new int[2]; 9 10 // search for left bound 11 start = 0; 12 end = A.length - 1; 13 while (start + 1 < end) { 14 mid = start + (end - start) / 2; 15 if (A[mid] == target) { 16 end = mid; 17 } else if (A[mid] < target) { 18 start = mid; 19 } else { 20 end = mid; 21 } 22 } 23 if (A[start] == target) { 24 bound[0] = start; 25 } else if (A[end] == target) { 26 bound[0] = end; 27 } else { 28 bound[0] = bound[1] = -1; 29 return bound; 30 } 31 32 // search for right bound 33 start = 0; 34 end = A.length - 1; 35 while (start + 1 < end) { 36 mid = start + (end - start) / 2; 37 if (A[mid] == target) { 38 start = mid; 39 } else if (A[mid] < target) { 40 start = mid; 41 } else { 42 end = mid; 43 } 44 } 45 if (A[end] == target) { 46 bound[1] = end; 47 } else if (A[start] == target) { 48 bound[1] = start; 49 } else { 50 bound[0] = bound[1] = -1; 51 return bound; 52 } 53 54 return bound; 55 } 56 }
我的解法:(第二次搜索从rst[0]开始,循环一直继续直到start和end都指向第一个tar)
(1.rst if(0); 2.lo,hi,while(<=)mid,if(>=)mid-1,else mid+1;if(=)mid; 3.if(-1)return; 4.lo,hi,while()mid,if(<=)mid+1,else mid-1;if(=)mid; 5.return)
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1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int[] rst = new int[]{-1,-1}; 4 if (A.length == 0) { 5 return rst; 6 } 7 int lo = 0, hi = A.length - 1; 8 while (lo <= hi) { 9 int mid = lo + (hi - lo) / 2; 10 if (A[mid] >= target) { 11 hi = mid - 1; 12 } else { 13 lo = mid + 1; 14 } 15 if (A[mid] == target) rst[0] = mid; 16 } 17 if (rst[0] == -1) { 18 return rst; 19 } 20 lo = rst[0]; 21 hi = A.length - 1; 22 while (lo <= hi) { 23 int mid = lo + (hi - lo) / 2; 24 if (A[mid] <= target) { 25 lo = mid + 1; 26 } else { 27 hi = mid - 1; 28 } 29 if (A[mid] == target) rst[1] = mid; 30 } 31 return rst; 32 } 33 }
类似:
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1 public int[] searchRange(int[] nums, int target) { 2 int[] result = new int[2]; 3 result[0] = findFirst(nums, target); 4 result[1] = findLast(nums, target); 5 return result; 6 } 7 8 private int findFirst(int[] nums, int target){ 9 int idx = -1; 10 int start = 0; 11 int end = nums.length - 1; 12 while(start <= end){ 13 int mid = (start + end) / 2; 14 if(nums[mid] >= target){ 15 end = mid - 1; 16 }else{ 17 start = mid + 1; 18 } 19 if(nums[mid] == target) idx = mid; 20 } 21 return idx; 22 } 23 24 private int findLast(int[] nums, int target){ 25 int idx = -1; 26 int start = 0; 27 int end = nums.length - 1; 28 while(start <= end){ 29 int mid = (start + end) / 2; 30 if(nums[mid] <= target){ 31 start = mid + 1; 32 }else{ 33 end = mid - 1; 34 } 35 if(nums[mid] == target) idx = mid; 36 } 37 return idx; 38 }
35. Search Insert Position: https://leetcode.com/problems/search-insert-position/
找出插入位置:
我的解法:(二分查找,=的时候作为pos即为mid,>的时候-1,< +1,最后的mid 如果大于等于tar,那么pos就是mid,小于tar,pos就是mid+1)
(1.lo,hi,rst,mid; 2.while(<=)if(=)return,if(>)-,if(<)+; 3.if([mid]>=)mid. else mid+1; 4.return)
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1 public class Solution { 2 public int searchInsert(int[] nums, int target) { 3 int lo = 0, hi = nums.length - 1; 4 int rst = 0; 5 int mid = 0; 6 while (lo <= hi) { 7 mid = lo + (hi - lo) / 2; 8 if (nums[mid] == target) { 9 return mid; 10 } else if (nums[mid] > target) { 11 hi = mid - 1; 12 } else { 13 lo = mid + 1; 14 } 15 } 16 if (nums[mid] >= target) { 17 rst = mid; 18 } else { 19 rst = mid + 1; 20 } 21 return rst; 22 } 23 }
类似:(return low; low=mid/low=mid+1)
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1 public int searchInsert(int[] A, int target) { 2 int low = 0, high = A.length-1; 3 while(low<=high){ 4 int mid = (low+high)/2; 5 if(A[mid] == target) return mid; 6 else if(A[mid] > target) high = mid-1; 7 else low = mid+1; 8 } 9 return low; 10 }
39. Combination Sum: https://leetcode.com/problems/combination-sum/
找出数组中所有相加可得到tar的组合(个数不限、使用次数不限):
解法:(回溯法; target-index作为新的target从i处继续判断; 判断语句放在for外,则是target和0比较;注意递归参数是(tar-[i]和i))(PS:逻辑简单,刚开始参数搞错,结果就一直不对;要学会根据错误结果推断原因)
(1.helper(nums,tar,rst,tmp,index) 2.if(0)rst.add,if(<)return; 3.for(index)add,helper(,tar-[i],rst,tmp,i),remove; 4.conbSum:sort,rst,tmp,helper,return)
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1 public class Solution { 2 public List<List<Integer>> combinationSum(int[] candidates, int target) { 3 List<List<Integer>> rst = new ArrayList<List<Integer>>(); 4 Arrays.sort(candidates); 5 List<Integer> tmp = new ArrayList<Integer>(); 6 helper(candidates, target, rst, tmp, 0); 7 return rst; 8 } 9 public void helper(int[] nums, int target, List<List<Integer>> rst, List<Integer> tmp, int index){ 10 if (target == 0) { 11 rst.add(new ArrayList<Integer>(tmp)); 12 return; 13 } 14 if (target < 0) { 15 return; 16 } 17 for (int i = index; i < nums.length; i++) { 18 tmp.add(nums[i]); 19 helper(nums, target - nums[i], rst, tmp, i); 20 tmp.remove(tmp.size() - 1); 21 } 22 } 23 }
40. Combination Sum II: https://leetcode.com/problems/combination-sum-ii/
含重复数组找出和为tar的组合(个数不限,只能使用一次):
我的解法:(与Combination Sum的解法相同; 区别:1.回溯时要把指针后移一位i+1; 2.遍历时要跳重,只能使用每轮遍历重复数字的第一个(i=index || [i]!=[i-1])
(1.helper(nums,tar,rst,tmp,index) 2.if(0)rst.add,if(<)return; 3.for(index) if(||&&),add helper(,tar-[i],rst,tmp,i+1),remove; 4.cSum:sort,rst,tmp,help,retu)
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1 public class Solution { 2 public List<List<Integer>> combinationSum2(int[] candidates, int target) { 3 Arrays.sort(candidates); 4 List<List<Integer>> rst = new ArrayList<>(); 5 List<Integer> tmp = new ArrayList<>(); 6 helper(candidates, target, rst, tmp, 0); 7 return rst; 8 } 9 public void helper(int[] nums, int target, List<List<Integer>> rst, List<Integer> tmp, int index) { 10 if (target == 0) { 11 rst.add(new ArrayList<Integer>(tmp)); 12 return; 13 } 14 if (target < 0) { 15 return; 16 } 17 for (int i = index; i < nums.length; i++) { 18 if (i == index || (i != index && nums[i] != nums[i - 1])) { 19 tmp.add(nums[i]); 20 helper(nums, target - nums[i], rst, tmp, i + 1); 21 tmp.remove(tmp.size() - 1); 22 } 23 } 24 } 25 }
48. Rotate Image: https://leetcode.com/problems/rotate-image/
顺时针九十度旋转二维矩阵: (todo)
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1 public class Solution { 2 public void rotate(int[][] matrix) { 3 for(int i = 0; i<matrix.length; i++){ 4 for(int j = i; j<matrix[0].length; j++){ 5 int temp = 0; 6 temp = matrix[i][j]; 7 matrix[i][j] = matrix[j][i]; 8 matrix[j][i] = temp; 9 } 10 } 11 for(int i =0 ; i<matrix.length; i++){ 12 for(int j = 0; j<matrix.length/2; j++){ 13 int temp = 0; 14 temp = matrix[i][j]; 15 matrix[i][j] = matrix[i][matrix.length-1-j]; 16 matrix[i][matrix.length-1-j] = temp; 17 } 18 } 19 } 20 }
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1 public class Solution { 2 public void rotate(int[][] matrix) { 3 if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { 4 return; 5 } 6 7 int length = matrix.length; 8 9 for (int i = 0; i < length / 2; i++) { 10 for (int j = 0; j < (length + 1) / 2; j++){ 11 int tmp = matrix[i][j]; 12 matrix[i][j] = matrix[length - j - 1][i]; 13 matrix[length -j - 1][i] = matrix[length - i - 1][length - j - 1]; 14 matrix[length - i - 1][length - j - 1] = matrix[j][length - i - 1]; 15 matrix[j][length - i - 1] = tmp; 16 } 17 } 18 } 19 }
53. Maximum Subarray: https://leetcode.com/problems/maximum-subarray/
连续子数组的最大和:
解法:(动态规划?)(A[i]前的 Sum<0, 那么就从A[i]作为新的起始点; Sum>0, 那么比较Sum+A[i]和max的大小)(很乱……)
(maxSoFar,maxEndingHere)
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1 public static int maxSubArray(int[] A) { 2 int maxSoFar=A[0], maxEndingHere=A[0]; 3 for (int i=1;i<A.length;++i){ 4 maxEndingHere= Math.max(maxEndingHere+A[i],A[i]); 5 maxSoFar=Math.max(maxSoFar, maxEndingHere); 6 } 7 return maxSoFar; 8 }
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1 public class Solution { 2 public int maxSubArray(int[] A) { 3 if (A == null || A.length == 0){ 4 return 0; 5 } 6 7 int max = Integer.MIN_VALUE, sum = 0; 8 for (int i = 0; i < A.length; i++) { 9 sum += A[i]; 10 max = Math.max(max, sum); 11 sum = Math.max(sum, 0); 12 } 13 14 return max; 15 } 16 }
54.Spiral Matrix: https://leetcode.com/problems/spiral-matrix/
返回矩阵的螺旋顺序:
解法:(画出矩阵图, 第一趟为[00]-[0n], [1n]-[mn], [m,n-1]-[m0], [m-1,0]-[1,0]; 第二趟,左右边界各加1,用count计数; 判断只剩一行或一列的break)
(1.rst,if(); 2.rows,cols,count; 3.while(*2<) for for if(==1) for for, count++; 4.return rst;)
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1 public class Solution { 2 public List<Integer> spiralOrder(int[][] matrix) { 3 List<Integer> rst = new ArrayList<>(); 4 if (matrix == null || matrix.length == 0) 5 return rst; 6 int rows = matrix.length; 7 int cols = matrix[0].length; 8 int count = 0; 9 while (count * 2 < rows && count * 2 < cols) { 10 for (int i = count; i < cols - count; i++) { 11 rst.add(matrix[count][i]); 12 } 13 for (int i = count + 1; i < rows - count; i++) { 14 rst.add(matrix[i][cols - count - 1]); 15 } 16 17 if (rows - 2 * count == 1 || cols - 2 * count == 1) 18 break; 19 20 for (int i = cols - count - 2; i >= count; i--) { 21 rst.add(matrix[rows - count - 1][i]); 22 } 23 for (int i = rows - count - 2; i >= count + 1; i--) { 24 rst.add(matrix[i][count]); 25 } 26 count++; 27 } 28 return rst; 29 } 30 }
55. Jump Game:https://leetcode.com/problems/jump-game/
步数不大于值从开始跳到结尾:
解法1:DP动态规划 (定义数组boolean[] can, 对于数组中任意i,true的条件是存在j<i满足can[j]=true且j+a[j]<=i;所以需要遍历i之前的所有值; O(n^2))(大数据TLE)
(1.if(); 2.[]can,[0]; 3.for(0,n-1)for(0,i) if(&&)[i]; 4.return [n-1];)
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1 public class Solution { 2 public boolean canJump(int[] nums) { 3 if (nums == null || nums.length == 0) { 4 return true; 5 } 6 boolean[] can = new boolean[nums.length]; 7 can[0] = true; 8 for (int i = 0; i < nums.length; i++) { 9 for (int j = 0; j < i; j++) { 10 if (can[j] && j + nums[j] >= i) { 11 can[i] = true; 12 } 13 } 14 } 15 return can[nums.length - 1]; 16 } 17 }
解法2:Greedy--根据已知条件得出局部最优解 (0-farthest-a[0];遍历数组,如果i在far范围内,那么far=max(far,a[i]+i); 最后判断far和n-1;)
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1 public class Solution { 2 public boolean canJump(int[] nums) { 3 if (nums == null || nums.length == 0) { 4 return true; 5 } 6 int farthest = nums[0]; 7 for (int i = 1; i < nums.length; i++) { 8 if (farthest >= i ) 9 farthest = Math.max(farthest, i + nums[i]); 10 } 11 return farthest >= nums.length - 1; 12 } 13 }
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1 public class Solution { 2 public boolean canJump(int[] A) { 3 // think it as merging n intervals 4 if (A == null || A.length == 0) { 5 return false; 6 } 7 int farthest = A[0]; 8 for (int i = 1; i < A.length; i++) { 9 if (i <= farthest && A[i] + i >= farthest) { 10 farthest = A[i] + i; 11 } 12 } 13 return farthest >= A.length - 1; 14 } 15 }
59. Spiral Matrix II: https://leetcode.com/problems/spiral-matrix-ii/
平方螺旋矩阵:
我的解法:(与Spiral Matrix解法相同, 区别是rows和cols都是n, 定义count计数四边已经铺了数字的边框,定义num++;对四边分别for循环赋值, 内部与最外圈的区别就是多了一个count,包括起始条件+/-count, 终止条件+/-count; 前两个for顺序遍历, 中间插入判断条件break, 后两个for倒序;)
(1.int[][],num,count; 2.while(*2<) for for if() for for; 3.return)
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1 public class Solution { 2 public int[][] generateMatrix(int n) { 3 int[][] Matrix = new int[n][n]; 4 int count = 0; 5 int num = 1; 6 while (count * 2 < n) { 7 for (int i = count; i < n - count; i++) { 8 Matrix[count][i] = num++; 9 } 10 for (int i = count + 1; i < n - count; i++) { 11 Matrix[i][n - count - 1] = num++; 12 } 13 14 if (count * 2 == n - 1) break; 15 16 for (int i = n - count - 2; i >= count; i--) { 17 Matrix[n - count - 1][i] = num++; 18 } 19 for (int i = n - count - 2; i >= count + 1; i--) { 20 Matrix[i][count] = num++; 21 } 22 count++; 23 } 24 return Matrix; 25 } 26 }
类似:(不定义count, 直接用n-2作为变化, 判断条件为n>0; 不过重复定义了xStart,yStart; 其实方法一样……)
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1 public class Solution { 2 public int[][] generateMatrix(int n) { 3 if (n < 0) { 4 return null; 5 } 6 7 int[][] result = new int[n][n]; 8 9 int xStart = 0; 10 int yStart = 0; 11 int num = 1; 12 13 while (n > 0) { 14 if (n == 1) { 15 result[yStart][xStart] = num++; 16 break; 17 } 18 19 for (int i = 0; i < n - 1; i++) { 20 result[yStart][xStart + i] = num++; 21 } 22 23 for (int i = 0; i < n - 1; i++) { 24 result[yStart + i][xStart + n - 1] = num++; 25 } 26 27 for (int i = 0; i < n - 1; i++) { 28 result[yStart + n - 1][xStart + n - 1 - i] = num++; 29 } 30 31 for (int i = 0; i < n - 1; i++) { 32 result[yStart + n - 1 - i][xStart] = num++; 33 } 34 35 xStart++; 36 yStart++; 37 n = n - 2; 38 } 39 40 return result; 41 } 42 }
62. Unique Paths: https://leetcode.com/problems/unique-paths/
网格首尾的所有可能路径:
Wrong!典型TLE解法:递归
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1 public class Solution { 2 public int uniquePaths(int m, int n) { 3 if (m == 1 || n == 1) return 1; 4 return (uniquePaths(m - 1, n) + uniquePaths(m, n - 1)); 5 } 6 }
解法:(for循环;第一列和第一行上的所有数sum为1,对于[i][j], 有递推关系sum=[i-1][j]+[i][j-1], return[m-1][n-1]) (Q: 所有递归均能改成for循环遍历+递推?)
(1.sum[][]; 2.for[i][0] for[0][i]; 3.for()for()[i][j]; 4.return[-1][-1])
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1 public class Solution { 2 public int uniquePaths(int m, int n) { 3 int[][] sum = new int[m][n]; 4 for (int i = 0; i < m ; i++) { 5 sum[i][0] = 1; 6 } 7 for (int i = 0; i < n; i++) { 8 sum[0][i] = 1; 9 } 10 for (int i = 1; i < m; i++) { 11 for (int j = 1; j < n; j++) { 12 sum[i][j] = sum[i - 1][j] + sum[i][j - 1]; 13 } 14 } 15 return sum[m - 1][n - 1]; 16 } 17 }
63. Unique Paths II: https://leetcode.com/problems/unique-paths-ii/
有障碍的网格的所有可能路径:
我的解法: (类似U Paths, 区别:对于第一行和第一列,如果在[i][0]出现1,那么i上的sum为1,i下的sum为零;用递推关系sum[i][0] = sum[i-1][0]来表示, 当存在a[i][0]为1时,则sum[i][0]=0,后面依然用递推关系……(这段似乎可以用指针优化,一旦发现第一个1,就把后面的sum都置零?); 其次 对于[i][j]区别就是多一个判断条件,如果a[i][j]=1,那么sum[i][j]=0)
(1.m,n,int,if,sum; 2.for(1,m)if(0)[i][0]=[i-1][0] else[i][0]=0; for(1,n); 3.for(1,m)(1,n)if(1)0 else[i][j]=[i-1][j]+[i][j-1]; 4.return)
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1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 int m = obstacleGrid.length; 4 int n = obstacleGrid[0].length; 5 int[][] sum = new int [m][n]; 6 if (obstacleGrid[0][0] == 1) return 0; 7 sum[0][0] = 1; 8 for (int i = 1; i < m; i++) { 9 if (obstacleGrid[i][0] == 0) 10 sum[i][0] = sum[i - 1][0]; 11 else sum[i][0] = 0; 12 } 13 for (int i = 1; i < n; i++) { 14 if (obstacleGrid[0][i] == 0) 15 sum[0][i] = sum[0][i - 1]; 16 else sum[0][i] = 0; 17 } 18 for (int i = 1; i < m; i++) { 19 for (int j = 1; j < n; j++) { 20 if (obstacleGrid[i][j] == 1) 21 sum[i][j] = 0; 22 else 23 sum[i][j] = sum[i -1][j] + sum[i][j - 1]; 24 } 25 } 26 return sum[m - 1][n - 1]; 27 } 28 }
类似: (第一行第一列遇到1后就break,其他基本相同)
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1 public class Solution { 2 public int uniquePathsWithObstacles(int[][] obstacleGrid) { 3 if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) { 4 return 0; 5 } 6 7 int n = obstacleGrid.length; 8 int m = obstacleGrid[0].length; 9 int[][] paths = new int[n][m]; 10 11 for (int i = 0; i < n; i++) { 12 if (obstacleGrid[i][0] != 1) { 13 paths[i][0] = 1; 14 } else { 15 break; 16 } 17 } 18 19 for (int i = 0; i < m; i++) { 20 if (obstacleGrid[0][i] != 1) { 21 paths[0][i] = 1; 22 } else { 23 break; 24 } 25 } 26 27 for (int i = 1; i < n; i++) { 28 for (int j = 1; j < m; j++) { 29 if (obstacleGrid[i][j] != 1) { 30 paths[i][j] = paths[i - 1][j] + paths[i][j - 1]; 31 } else { 32 paths[i][j] = 0; 33 } 34 } 35 } 36 37 return paths[n - 1][m - 1]; 38 } 39 }
64. Minimum Path Sum:https://leetcode.com/problems/minimum-path-sum/
网格路径上元素和的最小值:
我的解法:(类似U Paths 和UP II, 方法是动态规划/递推关系。对于grid[i][j], sum的值为sum[i-1][j]和sum[i][j-1]的小的值加上grid[i][j]; 其次对于第一行和第一列,sum就是[0][i-1]+grid[0][i]) (PS:Bug Free!! 一次AC!!)
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1 public class Solution { 2 public int minPathSum(int[][] grid) { 3 int m = grid.length; 4 int n = grid[0].length; 5 int[][] sum = new int[m][n]; 6 sum[0][0] = grid[0][0]; 7 for (int i = 1; i < m; i++) { 8 sum[i][0] = sum[i - 1][0] + grid[i][0]; 9 } 10 for (int i = 1; i < n; i++) { 11 sum[0][i] = sum[0][i - 1] + grid[0][i]; 12 } 13 for (int i = 1; i < m; i++) { 14 for (int j = 1; j < n; j++) { 15 sum[i][j] = Math.min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j]; 16 } 17 } 18 return sum[m - 1][n - 1]; 19 } 20 }
73. Set Matrix Zeroes:https://leetcode.com/problems/set-matrix-zeroes/
零所在的行和列都置零:
解法1:(O(m+n)space, 初始化row[]和col[], if(matrix[i][j])等于零, 就将row[i]和col[j]置一; 遍历row,将为1的i行置零; 同理col;)
(1.m,n,row,col; 2.for()for()if(=)1; 3.for()if()for()0; for()if()for()0; )
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1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 int m = matrix.length; 4 int n = matrix[0].length; 5 int[] row = new int[m]; 6 int[] col = new int[n]; 7 for (int i = 0; i < m; i++) { 8 for (int j = 0; j < n; j++) { 9 if (matrix[i][j] == 0) { 10 row[i] = 1; 11 col[j] = 1; 12 } 13 } 14 } 15 for (int i = 0; i < m; i++) { 16 if (row[i] == 1) { 17 for (int j = 0; j < n; j++) { 18 matrix[i][j] = 0; 19 } 20 } 21 } 22 for (int j = 0; j < n; j++) { 23 if (col[j] == 1) { 24 for (int i = 0; i < m; i++) { 25 matrix[i][j] = 0; 26 } 27 } 28 } 29 30 } 31 }
解法2:(In place 用第一行和第一列作为标记; 从[1][1]后遍历, 如果[i][j]=0, 那么就标记对应的[i][0]和[0][j]=0; 但需要单独考虑row0和col0; 如果存在i=0或j=0时[i][j]=0,需要先暂时用fr和fc标记, 最后额外将first row或者first col置零)
(1.m,n,fr,fc; 2.for()for()if(=) if()if() 0 0; 3.for(1)for(1)if(||)0; 4.if(fr) if(fc);)
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1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 int m = matrix.length; 4 int n = matrix[0].length; 5 boolean fr = false; 6 boolean fc = false; 7 for (int i = 0; i < m; i++) { 8 for (int j = 0; j < n; j++) { 9 if (matrix[i][j] == 0) { 10 if (i == 0) fr = true; 11 if (j == 0) fc = true; 12 matrix[i][0] = 0; 13 matrix[0][j] = 0; 14 } 15 } 16 } 17 for (int i = 1; i < m; i++) { 18 for (int j = 1; j < n; j++) { 19 if (matrix[i][0] == 0 || matrix[0][j] == 0) { 20 matrix[i][j] = 0; 21 } 22 } 23 } 24 if (fr) { 25 for (int i = 0; i < n; i++) { 26 matrix[0][i] = 0; 27 } 28 } 29 if (fc) { 30 for (int i = 0; i < m; i++) { 31 matrix[i][0] = 0; 32 } 33 } 34 } 35 }
74. Search a 2D Matrix: https://leetcode.com/problems/search-a-2d-matrix/
查找二维矩阵: (参考LintCode 1-30 里的T28)
解法1: 暴力解法 (for)
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1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 int row = matrix.length; 4 int col = matrix[0].length; 5 for (int i = 0; i < row; i++) { 6 for (int j = 0; j < col; j++) { 7 if (matrix[i][j] == target) 8 return true; 9 } 10 } 11 return false; 12 } 13 }
解法2: 两次二分(第一次找到tarRow,防止遗漏while(+1<)+mid+if判断; 第二次while(<=)+ mid+1/-1;)
(1.ifif; 2.start,end,row; 3.while(+1<)mid if(=)if(</>)mid; 4.if(<=)elseif(<=)else false; 5.while(<=)if eif e; 6.return false;)
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public class Solution { /** * @param matrix, a list of lists of integers * @param target, an integer * @return a boolean, indicate whether matrix contains target */ public boolean searchMatrix(int[][] matrix, int target) { // write your code here if (matrix == null || matrix.length == 0){ return false; } if (matrix[0] == null || matrix[0].length == 0){ return false; } int start = 0; int end = matrix.length - 1; int row ; while (start + 1 < end) { int mid = start + (end - start)/2; if (matrix[mid][0] == target){ return true; } else if (matrix[mid][0] < target) { start = mid ; } else { end = mid ; } } if (matrix[end][0] <= target){ row = end; } else if (matrix[start][0] <= target){ row = start; } else { return false; } start = 0; end = matrix[0].length - 1; while (start <= end) { int mid = start + (end - start)/2; if (matrix[row][mid] == target){ return true; } else if (matrix[row][mid] < target) { start = mid + 1; } else { end = mid - 1; } } return false; } }
解法3:一次二分(把matrix看成数组, m*n-1; row=mid/col, col=mid%col;)
(1.ifif; 2.row,col,start,end; 3.while(<=)mid num=[/][%], if()e if() e; 4.return false; )
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1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 if (matrix == null || matrix.length == 0) 4 return false; 5 if (matrix[0] == null || matrix[0].length == 0) 6 return false; 7 int row = matrix.length; 8 int col = matrix[0].length; 9 10 int start = 0, end = row * col - 1; 11 while (start <= end) { 12 int mid = start + (end - start) / 2; 13 if (matrix[mid / col][mid % col] == target) { 14 return true; 15 } else if (matrix[mid / col][mid % col] > target) { 16 end = mid - 1; 17 } else 18 start = mid + 1; 19 } 20 return false; 21 } 22 }
75. Sort Colors: https://leetcode.com/problems/sort-colors/
只含0,1,2的数组排序:
解法1:计数 两趟for循环:
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1 public class Solution { 2 public void sortColors(int[] nums) { 3 int cnt0 = 0, cnt1 = 0; 4 for (int i = 0; i < nums.length; i++) { 5 if (nums[i] == 0) 6 cnt0++; 7 else if (nums[i] == 1) 8 cnt1++; 9 } 10 for (int i = 0; i < nums.length; i++) { 11 if (i < cnt0) 12 nums[i] = 0; 13 else if (i < cnt0 + cnt1) 14 nums[i] = 1; 15 else nums[i] = 2; 16 } 17 } 18 }
解法2:Two Pointers (左指针指向0的后一位, 右指针指向2的前一位, i遍历l; 如果遍历到[i]为0,交换pl和i,pr++,i++; 如果遍历到2,交换pr和i,pr--; 遍历到1,i++;)(注意:循环条件i<=pr; [i]=2时,swap后i不增;)
(1.pl,pr,i; 2.for(<=)if()swap(++,++), if()++, if()swap(--); 3.swap)
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1 public class Solution { 2 public void sortColors(int[] nums) { 3 int pl = 0, pr = nums.length - 1; 4 for (int i = 0; i <= pr;) { 5 if (nums[i] == 0) 6 swap(nums, i++, pl++); 7 else if (nums[i] == 1) 8 i++; 9 else 10 swap(nums, i, pr--); 11 } 12 } 13 public void swap (int[]a, int i, int j) { 14 int tmp = a[i]; 15 a[i] = a[j]; 16 a[j] = tmp; 17 } 18 }
78. Subsets: https://leetcode.com/problems/subsets/
数组的子集数组:(参考LintCode 1-30: T17)
解法1:(1.rst,if(); 2.tmp,helper,return; 3.helper(nums,rst,tmp,index), add, for(index) add,help(i+1),remove)
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1 public class Solution { 2 public List<List<Integer>> subsets(int[] nums) { 3 ArrayList<List<Integer>> rst = new ArrayList<List<Integer>>(); 4 if (nums.length == 0) { 5 rst.add(new ArrayList<Integer>()); 6 return rst; 7 } 8 List<Integer> tmp = new ArrayList<Integer>(); 9 helper(nums, rst, tmp, 0); 10 return rst; 11 } 12 public void helper(int[] nums, List<List<Integer>> rst, List<Integer> tmp, int index){ 13 rst.add(new ArrayList<Integer>(tmp)); 14 15 for (int i = index; i < nums.length; i++) { 16 tmp.add(nums[i]); 17 helper(nums, rst, tmp, i + 1); 18 tmp.remove(tmp.size() - 1); 19 } 20 } 21 }
解法2:位运算(用bit位来表示这一位的number要不要取,第一位有1,0即取和不取2种可能性。所以只要把0到N种可能都用bit位表示,再把它转化为数字集合,就可以了)
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1 class Solution { 2 /** 3 * @param S: A set of numbers. 4 * @return: A list of lists. All valid subsets. 5 */ 6 public ArrayList<ArrayList<Integer>> subsets(int[] nums) { 7 ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); 8 int n = nums.length; 9 Arrays.sort(nums); 10 11 // 1 << n is 2^n 12 // each subset equals to an binary integer between 0 .. 2^n - 1 13 // 0 -> 000 -> [] 14 // 1 -> 001 -> [1] 15 // 2 -> 010 -> [2] 16 // .. 17 // 7 -> 111 -> [1,2,3] 18 for (int i = 0; i < (1 << n); i++) { 19 ArrayList<Integer> subset = new ArrayList<Integer>(); 20 for (int j = 0; j < n; j++) { 21 // check whether the jth digit in i's binary representation is 1 22 if ((i & (1 << j)) != 0) { 23 subset.add(nums[j]); 24 } 25 } 26 result.add(subset); 27 } 28 29 return result; 30 } 31 }
79. Word Search:https://leetcode.com/problems/word-search/
二维矩阵查找字符串(todo)
80. Remove Duplicates from Sorted Array II: https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
解法:<Two Pointers> (index指向"新"数组的后一位, i指向原数组; 如果i所指的数num和index-2所指的数相同, 表面new Array里已经包含两个数了, 那么就跳过这个数;)
(1.index; 2.for() if(<2||!=[index-2])= 3.return index;)
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1 public class Solution { 2 public int removeDuplicates(int[] nums) { 3 int index = 0; 4 for (int i = 0; i < nums.length; i++) { 5 if (i < 2 || nums[i] != nums[index - 2]) { 6 nums[index++] = nums[i]; 7 } 8 } 9 return index; 10 } 11 }
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1 Remove Duplicates from Sorted Array(no duplicates) : 2 3 public int removeDuplicates(int[] nums) { 4 int i = 0; 5 for(int n : nums) 6 if(i < 1 || n > nums[i - 1]) 7 nums[i++] = n; 8 return i; 9 } 10 Remove Duplicates from Sorted Array II (allow duplicates up to 2): 11 12 public int removeDuplicates(int[] nums) { 13 int i = 0; 14 for (int n : nums) 15 if (i < 2 || n > nums[i - 2]) 16 nums[i++] = n; 17 return i; 18 }
81. Search in Rotated Sorted Array II: https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
(todo)
90. Subsets II: https://leetcode.com/problems/subsets-ii/
含重复元素的子集:
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1 public class Solution { 2 public List<List<Integer>> subsetsWithDup(int[] nums) { 3 List<List<Integer>> rst = new ArrayList<>(); 4 if (nums.length == 0) { 5 rst.add(new ArrayList<Integer>()); 6 return rst; 7 } 8 Arrays.sort(nums); 9 List<Integer> tmp = new ArrayList<Integer>(); 10 helper(nums, rst, tmp, 0); 11 return rst; 12 } 13 public void helper(int[] nums, List<List<Integer>> rst, List<Integer> tmp, int index) { 14 rst.add(new ArrayList<Integer>(tmp)); 15 for (int i = index; i < nums.length; i++) { 16 if (i == index || (i > index && nums[i] != nums[i - 1])){ 17 tmp.add(nums[i]); 18 helper(nums, rst, tmp, i + 1); 19 tmp.remove(tmp.size() - 1); 20 } 21 22 } 23 } 24 }