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;
        
    }
}