376. Wiggle Subsequence 类别:动态规划 难度:medium

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匿名技术用户   2021-1-8 06:09   31   0

题目:

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

程序:

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


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