【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][0] = max(dp[k][1] + 1, dp[i][0]);
dp[i][1] = max(dp[k][0] + 1, dp[i][1]);
dp[i][0] represent the longest wanted subsequence with a positive sum ending;
dp[i][1] 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 == 0) return 0;int dp[n][2];for(int i = 0; i < n; i ++){dp[i][0] = dp[i][1] = 0;}int ans = 0;for(int i = 1; i < n; i ++){for(int k = 0; k < i; k ++){if(nums[i] > nums[k]){dp[i][0] = max(dp[i][0], dp[k][1] + 1);}else if(nums[i] < nums[k]){dp[i][1] = max(dp[i][1], dp[k][0] + 1);}}ans = max(dp[i][0], dp[i][1]);}return ans + 1;}
};

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

class Solution {
public:int wiggleMaxLength(vector<int>& nums) {const int n = nums.size();if(n == 0) return 0;int dp[n][2];//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[0][0] = dp[0][1] = 1;for(int i = 1; i < n; i ++){if(nums[i] > nums[i - 1]){dp[i][0] = dp[i - 1][1] + 1;dp[i][1] = dp[i - 1][1];}else if(nums[i] < nums[i -1]){dp[i][1] = dp[i - 1][0] + 1;dp[i][0] = dp[i - 1][0];}else{dp[i][0] = dp[i - 1][0];dp[i][1] = dp[i - 1][1];}} return max(dp[n - 1][0], dp[n - 1][1]);}
};

转载于:https://www.cnblogs.com/luntai/p/5899048.html

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