leetcode-376-最长摆动序列

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本文介绍了一种通过动态规划求解最长摆动子序列的方法。该方法利用两个数组分别记录以正斜率和负斜率结束的最长摆动序列长度，并通过遍历输入序列更新这两个数组，最终返回两数组中的最大值作为最长摆动子序列的长度。

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

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        //动态规划 两个数组，每个存到目前为止的数目
        //用up[i]和down[i]分别记录到第i个元素为止 以上升沿和下降沿[结束] 的最长“摆动”序列长度
        int n = nums.size();
        if(n==0 || n==1) return n;
        vector<int> up(n);
        vector<int> down(n);
        up[0] = down[0] = 1;
        for(int i=1;i<n;i++){
            if(nums[i] < nums[i-1]){
                down[i] = up[i-1] + 1;
                up[i] = up[i-1];
            }else if(nums[i] > nums[i-1]){
                up[i] = down[i-1] + 1;
                down[i] = down[i-1];
            }else{
                up[i] = up[i-1];
                down[i] = down[i-1];
            }
        }
        return max(up[n-1], down[n-1]);
    }
};