LeetCode—376. Wiggle Subsequence
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Wiggle Subsequence思路:贪心算法,每次只要跟前一个比,如果符合条件,则获取,如果不符合条件,说明提升了可以比较的空间
GitHub地址:https://github.com/corpsepiges/leetcode
目前java版本的答案大约进度是免费的差40题,python大约是一半,其他的等以后再补充。
点此进入如果可以的话,请点一下star,谢谢。
<span style="font-size:14px;">public class Solution { public int wiggleMaxLength(int[] nums) { if (nums.length==0) { return 0; } Boolean flag=null; int ans=1; for (int i = 1; i < nums.length; i++) { if (nums[i]!=nums[i-1]&&(flag==null||flag==nums[i]>nums[i-1])) { flag=nums[i]<nums[i-1]; ans++; } } return ans; }}</span>
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