【Leetcode】376. Wiggle Subsequence

Description：　　

A sequence of numbers is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a wiggle sequence.

For example, [1,7,4,9,2,5] is a wiggle sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast,[1,4,7,2,5] and [1,7,4,5,5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

Examples:

Input: [1,7,4,9,2,5]
Output: 6
The entire sequence is a wiggle sequence.

Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

Input: [1,2,3,4,5,6,7,8,9]
Output: 2

　　I got this problem by mocking which was given 40 mins. However failed,  WTF! At the begining, I concluded it was an dp problem. I was stuck in how to solve it in one loop n(O(n) timespace). then I try to figure out the trans-fomula:

dp[i][] = max(dp[k][] + , dp[i][]);
dp[i][] = max(dp[k][] + , dp[i][]);
dp[i][] represent the longest wanted subsequence with a positive sum ending;
dp[i][] similarly but with a negative sum ending;

　　You must solve it in time which may sacrifice the timespace!

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        const int n = nums.size();
        if(n == ) return ;
        int dp[n][];
        for(int i = ; i < n; i ++){
            dp[i][] = dp[i][] = ;
        }int ans = ;
        for(int i = ; i < n; i ++){
            for(int k = ; k < i; k ++){
               if(nums[i] > nums[k]){
                   dp[i][] = max(dp[i][], dp[k][] + );
               }else if(nums[i] < nums[k]){
                   dp[i][] = max(dp[i][], dp[k][] + );
               }
            }
            ans = max(dp[i][], dp[i][]);
        }
        return ans + ;
    }
};

　　Finally, I optimize the solution to O(n).

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        const int n = nums.size();
        if(n == ) return ;
        int dp[n][];
        //dp[i][0] : Before i the longest wanted subsequence ending with a positive ending.
        //dp[i][1] : Before i the longest wanted subsequence ending with a negative ending.
        dp[][] = dp[][] = ;
        for(int i = ; i < n; i ++){
            if(nums[i] > nums[i - ]){
                dp[i][] = dp[i - ][] + ;
                dp[i][] = dp[i - ][];
            }else if(nums[i] < nums[i -]){
                dp[i][] = dp[i - ][] + ;
                dp[i][] = dp[i - ][];
            }else{
                dp[i][] = dp[i - ][];
                dp[i][] = dp[i - ][];
            }
        }
        return max(dp[n - ][], dp[n - ][]);
    }
};