LeetCode "Wiggle Subsequence" !

Another interesting DP. Lesson learnt: how you define state is crucial..

1. if DP[i] is defined as, longest wiggle(up\down) subseq AT number i, you will have O(n^2) solution

class Solution
{
    struct Rec
    {
        Rec(): mlen_dw(), mlen_up(){}
        Rec(int ldw, int lup): mlen_dw(ldw), mlen_up(lup){}
        int mlen_dw;
        int mlen_up;
    };
public:
    int wiggleMaxLength(vector<int>& nums)
    {
        int n = nums.size();
        if(n < ) return n;

        vector<Rec> dp(n);
        dp[].mlen_up = dp[].mlen_dw = ;

        int ret = ;
        for(int i = ; i < n; i ++)
        {
            int cv = nums[i];
            for(int j = i - ; j >= max(, ret - ); j --)
            {
                if(cv > nums[j])
                {
                    dp[i].mlen_up = max(dp[i].mlen_up, dp[j].mlen_dw + );
                }
                else if(cv < nums[j])
                {
                    dp[i].mlen_dw = max(dp[i].mlen_dw, dp[j].mlen_up + );
                }
                ret = max(ret, max(dp[i].mlen_dw, dp[i].mlen_up));
            }
        }

        return ret;
    }
};

2. if DP[i] is defined as, longest wiggle(up\down) subseq SO FAR UNTIL number i, you will have O(n) solution

class Solution
{
    struct Rec
    {
        Rec(): mlen_dw(), mlen_up(){}
        Rec(int ldw, int lup): mlen_dw(ldw), mlen_up(lup){}
        int mlen_dw;
        int mlen_up;
    };
public:
    int wiggleMaxLength(vector<int>& nums)
    {
        int n = nums.size();
        if(n < ) return n;

        vector<Rec> dp(n);
        dp[].mlen_up = dp[].mlen_dw = ;

        int ret = ;
        for(int i = ; i < n; i ++)
        {
            int cv = nums[i];
            dp[i] = dp[i - ];
            if(cv > nums[i - ])
            {
                dp[i].mlen_up = max(dp[i].mlen_up, dp[i - ].mlen_dw + );
            }
            else if(cv < nums[i - ])
            {
                dp[i].mlen_dw = max(dp[i].mlen_dw, dp[i - ].mlen_up + );
            }
            ret = max(ret, max(dp[i].mlen_dw, dp[i].mlen_up));
        }

        return ret;
    }
};

3. And, there's always smarter solution - GREEDY!
https://discuss.leetcode.com/topic/52074/concise-10-lines-code-0ms-acepted