題目

https://leetcode-cn.com/problems/longest-harmonious-subsequence/

題解

我的解法

測試用例

[1]
[1,2]
[2,1]
[3,1]
[3,1,1]
[3,5,5]
[1,3,5,7]
[2,2]
[1,2,1,2,1,2,1,2,1,2]
[3,3,3,2,2,2,1,1,1]
[3,3,2]
[1,3,2,2,5,2,3,7]

import java.util.Arrays;class Solution {public int findLHS(int[] nums) {if (nums.length == 1) return 0;Arrays.sort(nums);int maxSum = 0;int numPre = Integer.MAX_VALUE;int cntPre = 0;int cntCur = 1;for (int i = 0; i < nums.length - 1; i++) {if (nums[i] == nums[i + 1]) {cntCur++;} else {if (numPre + 1 == nums[i]) { // update maxSum only when it's continuousmaxSum = Math.max(maxSum, cntPre + cntCur);}cntPre = cntCur;cntCur = 1;numPre = nums[i];}}// boundary: the last numberif (nums[nums.length - 2] != nums[nums.length - 1]) {cntPre = cntCur;cntCur = 1;}if (numPre + 1 == nums[nums.length - 1]) { // update maxSum only when it's continuousmaxSum = Math.max(maxSum, cntPre + cntCur);}return maxSum;} }

評論區優雅解法

排序后，用兩個指針，類似于窗口

class Solution {public int findLHS(int[] nums) {Arrays.sort(nums);int begin = 0,res = 0;for(int end = 0;end < nums.length;end++){while(nums[end] - nums[begin] > 1)begin++;if(nums[end] - nums[begin] == 1)res = Math.max(res,end - begin + 1);}return res;} }

總結

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