34.Find First and Last Position of Element in Sorted Array---头条面试题、《剑指offer》38
题目大意:找出一串升序数组中target值的起始下标和结束下标值,如果不存在则返回{-1,-1}。
解法一:用二分查找,找到数组中的target,然后找其左边和右边的target下标值。代码如下(耗时11ms):
1 public int[] searchRange(int[] nums, int target) { 2 if(nums == null || nums.length == 0) { 3 int[] r = {-1, -1}; 4 return r; 5 } 6 int low = 0, high = nums.length - 1; 7 int start = -1, end = -1; 8 while(low <= high) { 9 int mid = (low + high) / 2; 10 if(nums[mid] < target) { 11 low = mid + 1; 12 } 13 else if(nums[mid] > target) { 14 high = mid - 1; 15 } 16 else { 17 //找左边起始下标 18 for(int i = mid; i >= 0; i--) { 19 if(nums[i] == target) { 20 start = i; 21 } 22 else { 23 break; 24 } 25 } 26 //找右边终止下标 27 for(int i = mid; i < nums.length; i++) { 28 if(nums[i] == target) { 29 end = i; 30 } 31 else { 32 break; 33 } 34 } 35 break; 36 } 37 } 38 int[] res = {start, end}; 39 return res; 40 }
解法二:直接暴力,一次遍历,找重复值。代码如下(耗时10ms):
1 public int[] searchRange(int[] nums, int target) { 2 int start = -1, end = -1; 3 boolean mark = false; 4 for(int i = 0; i < nums.length; i++) { 5 if(nums[i] == target) { 6 if(mark == false) { 7 start = end = i; 8 mark = true; 9 } 10 else { 11 end = i; 12 } 13 } 14 } 15 int[] res = {start, end}; 16 return res; 17 }
解法三:真正的二分查找。法一其实复杂度还是o(n)。应该先对起始下标进行二分查找,然后再对结束下标进行二分查找。代码如下(耗时5ms):
1 public int[] searchRange(int[] nums, int target) { 2 int left = 0, right = nums.length - 1; 3 int[] res = {-1, -1}; 4 if(nums.length == 0) { 5 return res; 6 } 7 //二分找到起始下标 8 while(left < right) { 9 int mid = (left + right) / 2; 10 //这里比较左值,如果<,则left更新,否则left不会更新 11 //所以left不更新有两种情况:>或= 12 if(nums[mid] < target) { 13 left = mid + 1; 14 } 15 //这里统统修改右值 16 else { 17 right = mid; 18 } 19 } 20 if(nums[left] != target) { 21 return res; 22 } 23 res[0] = left; 24 left = 0; 25 right = nums.length - 1; 26 //二分找到结束下标 27 while(left < right) { 28 //这里要+1,否则会出错 29 int mid = (left + right) / 2 + 1; 30 //比较右值,如果>,则right更新 31 if(nums[mid] > target) { 32 right = mid - 1; 33 } 34 //修改Left 35 else { 36 left = mid; 37 } 38 } 39 res[1] = right; 40 return res; 41 }