标签：sub 情况 style 序列 res public 长度 更新 pre
用dp解
1）up定义为nums[i-1] < nums[i]
down nums[i-1] > nums[i]
两个dp数组,
up[i],记录包含nums[i]且nums[i-1] < nums[i]的最长子序列长度
down[], 记录包含nums[i]nums[i-1] > nums[i]的最长子序列长度
2）更新方程
有三种情况 nums[i-1] <or=or> nums[i]
a) up[i] = down[i-1] + 1;
down[i] = down[i-1]
b) up[i] = up[i-1]
down[i] = down[i-1]
c) up[i] = up[i-1]
down[i] = up[i-1] + 1;
1 class Solution { 2 public int wiggleMaxLength(int[] nums) { 3 if(nums == null || nums.length == 0) 4 return 0; 5 int len = nums.length; 6 int[] up = new int[len]; 7 int[] down = new int[len]; 8 int res = 1; 9 10 up[0] = 1; 11 down[0] = 1; 12 13 for(int i=1; i<len; i++){ 14 if(nums[i] > nums[i-1]){ 15 up[i] = down[i-1] + 1; 16 down[i] = down[i-1]; 17 }else if(nums[i] < nums[i-1]){ 18 up[i] = up[i-1]; 19 down[i] = up[i-1] + 1; 20 }else{ 21 up[i] = up[i-1]; 22 down[i] = down[i-1]; 23 } 24 25 res = Math.max(res, Math.max(up[i], down[i])); 26 } 27 28 return res; 29 30 } 31 }
leetcode 376Wiggle Subsequence
标签：sub 情况 style 序列 res public 长度 更新 pre
原文地址：https://www.cnblogs.com/hwd9654/p/10926101.html
