标签：
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(0), mlen_up(0){} 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 < 2) return n; vector<Rec> dp(n); dp[0].mlen_up = dp[0].mlen_dw = 1; int ret = 1; for(int i = 1; i < n; i ++) { int cv = nums[i]; for(int j = i - 1; j >= max(0, ret - 2); j --) { if(cv > nums[j]) { dp[i].mlen_up = max(dp[i].mlen_up, dp[j].mlen_dw + 1); } else if(cv < nums[j]) { dp[i].mlen_dw = max(dp[i].mlen_dw, dp[j].mlen_up + 1); } 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(0), mlen_up(0){} 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 < 2) return n; vector<Rec> dp(n); dp[0].mlen_up = dp[0].mlen_dw = 1; int ret = 1; for(int i = 1; i < n; i ++) { int cv = nums[i]; dp[i] = dp[i - 1]; if(cv > nums[i - 1]) { dp[i].mlen_up = max(dp[i].mlen_up, dp[i - 1].mlen_dw + 1); } else if(cv < nums[i - 1]) { dp[i].mlen_dw = max(dp[i].mlen_dw, dp[i - 1].mlen_up + 1); } 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
LeetCode "Wiggle Subsequence" !
标签：
原文地址：http://www.cnblogs.com/tonix/p/5700042.html
