LeetCode 697. Degree of an Array
Given a non-empty array of non-negative integers nums
, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums
, that has the same degree as nums
.
Example 1:
Input: [1, 2, 2, 3, 1] Output: 2 Explanation: The input array has a degree of 2 because both elements 1 and 2 appear twice. Of the subarrays that have the same degree: [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2] The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2] Output: 6
Note:
nums.length
will be between 1 and 50,000.nums[i]
will be an integer between 0 and 49,999.
first:
class Solution {
public int findShortestSubArray(int[] nums) {
int[] counts = findCounts(nums);
int degree = 0;
for(int i=0; i<counts.length; i++) {
degree = Math.max(degree, counts[i]);
}//find degree
ArrayList<Integer> degreeNums = new ArrayList<Integer>();
for(int i=0; i<counts.length; i++) {
if(counts[i] == degree) {
degreeNums.add(i);
}
}//find all degreeNums
int minSubLength = 0;
for(Integer degreeNum : degreeNums) {
int count = 0, start = 0, end = 0;
for(int i=0; i<nums.length; i++) {
if(nums[i] == degreeNum) {
//start = i;
count++;
}
if(nums[i] == degreeNum && count == 0) {
start = i;
//count++;
}
if(nums[i] == degreeNum && count == counts[i] - 1) {
end = i;
}
}
minSubLength = Math.min(minSubLength, end - start + 1);
}//find subLength
return minSubLength;
}
public int[] findCounts(int[] nums) {
int[] counts = new int[50000];
for(int i=0; i<nums.length; i++) {
counts[nums[i]]++;
}
return counts;
}
}
分析:
java.lang.Math与java.math包不同。
result:
Submission Result: Wrong Answer Input: [1,2,2,3,1] Output: 0 Expected: 2
分析:
second:
class Solution { public int findShortestSubArray(int[] nums) { int[] counts = findCounts(nums); int degree = 0; for(int i=0; i<counts.length; i++) { degree = Math.max(degree, counts[i]); }//find degree ArrayList<Integer> degreeNums = new ArrayList<Integer>(); for(int i=0; i<counts.length; i++) { if(counts[i] == degree) { degreeNums.add(i); } }//find all degreeNums int minSubLength = 0; for(Integer degreeNum : degreeNums) { int count = 0, start = 0, end = 0; for(int i=0; i<nums.length; i++) { if(nums[i] == degreeNum) { //start = i; count++; } if(nums[i] == degreeNum && count == 0) { start = i; //count++; } if(nums[i] == degreeNum && count == counts[i] - 1) { end = i; } } if(minSubLength == 0) { minSubLength = Math.max(minSubLength, end - start + 1); } else { minSubLength = Math.min(minSubLength, end - start + 1); } }//find subLength return minSubLength; } public int[] findCounts(int[] nums) { int[] counts = new int[50000]; for(int i=0; i<nums.length; i++) { counts[nums[i]]++; } return counts; } }
结果:
Submission Result: Wrong Answer Input: [1,2,2,3,1] Output: 1 Expected: 2
third:
class Solution { public int findShortestSubArray(int[] nums) { int[] counts = findCounts(nums); int degree = 0; for(int i=0; i<counts.length; i++) { degree = Math.max(degree, counts[i]); }//find degree ArrayList<Integer> degreeNums = new ArrayList<Integer>(); for(int i=0; i<counts.length; i++) { if(counts[i] == degree) { degreeNums.add(i); } }//find all degreeNums int minSubLength = 0; for(Integer degreeNum : degreeNums) { int count = 0, start = 0, end = 0; for(int i=0; i<nums.length; i++) { /*if(nums[i] == degreeNum && count == 0) { start = i; //count++; } if(nums[i] == degreeNum) { //start = i; count++; } if(nums[i] == degreeNum && count == counts[i] - 1) { end = i; }*/ if(nums[i] == degreeNum) { if(count == 0) { start = i; } count++; if(count == counts[i]) { end = i; } } } if(minSubLength == 0) { minSubLength = Math.max(minSubLength, end - start + 1); } else { minSubLength = Math.min(minSubLength, end - start + 1); } }//find subLength return minSubLength; } public int[] findCounts(int[] nums) { int[] counts = new int[50000]; for(int i=0; i<nums.length; i++) { counts[nums[i]]++; } return counts; } }
分析:
这一步花了好长时间。 代码一多就容易出错,而且难定位。最后发现是 if(count == counts[i]) {中的数组下标出了问题。
fourth:
class Solution { public int findShortestSubArray(int[] nums) { int[] counts = findCounts(nums); int degree = 0; for(int i=0; i<counts.length; i++) { degree = Math.max(degree, counts[i]); }//find degree System.out.println(degree); ArrayList<Integer> degreeNums = new ArrayList<Integer>(); for(int i=0; i<counts.length; i++) { if(counts[i] == degree) { degreeNums.add(i); } }//find all degreeNums System.out.println(degreeNums); int minSubLength = 0; for(Integer degreeNum : degreeNums) { int count = 0, start = 0, end = 0; for(int i=0; i<nums.length; i++) { if(nums[i] == degreeNum) { if(count == 0) { start = i; } count++; if(count == counts[degreeNum]) { end = i; } } } if(minSubLength == 0) { minSubLength = Math.max(minSubLength, end - start + 1); } else { minSubLength = Math.min(minSubLength, end - start + 1); } }//find subLength return minSubLength; } public int[] findCounts(int[] nums) { int[] counts = new int[50000]; for(int i=0; i<nums.length; i++) { counts[nums[i]]++; } return counts; } }
结果:
分析:
时间效率不是很好,但懒得再优化了。
总结:
分类:
LeetCode
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