leetcode 376Wiggle Subsequence

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;

 class Solution {
     public int wiggleMaxLength(int[] nums) {
         if(nums == null || nums.length == 0)
             return 0;
         int len = nums.length;
         int[] up = new int[len];
         int[] down = new int[len];
         int res = 1;
 
         up[0] = 1;
         down[0] = 1;
 
         for(int i=1; i<len; i++){
             if(nums[i] > nums[i-1]){
                 up[i] = down[i-1] + 1;
                 down[i] = down[i-1];
             }else if(nums[i] < nums[i-1]){
                 up[i] = up[i-1];
                 down[i] = up[i-1] + 1;
             }else{
                 up[i] = up[i-1];
                 down[i] = down[i-1];
             }
 
             res = Math.max(res, Math.max(up[i], down[i]));
         }
 
         return res;
 
     }
 }