LeetCode 376. Wiggle Subsequence 题解

最新推荐文章于 2025-03-10 16:35:06 发布

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最新推荐文章于 2025-03-10 16:35:06 发布

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本文介绍了一种寻找给定整数序列中最长摆动子序列的方法。摆动序列是指序列中相邻元素差值严格交替正负的序列。文章详细解释了如何通过检查相邻元素之间的差异来确定序列是否为摆动序列，并给出了实现这一功能的具体算法。

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【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

【题解】

首先，用数组difference记录原来数组每相邻两个数的差，差若为0不保存；

然后，检查difference中元素的个数，若为0，表示原数组每个元素均相等，那么结果为1；

接着，用一个临时变量nTmp记录difference当前的元素difference[i]，若后一个元素difference[j] （j > i），若nTmp与difference[j]的符号相反，则长度加1，同时nTmp赋值为difference[j] ；

最后，将得到的长度加1即可得到最后结果

【代码】

    int wiggleMaxLength(vector<int>& nums) {
        int nSize = nums.size();
        if (nSize < 2)
            return nSize;
            
        vector<int> difference;
        for (int i = 1; i < nSize; i++) {
            int nDiff = nums[i] - nums[i - 1];
            if (nDiff != 0)
                difference.push_back(nDiff);
        }
        int nSize_diff = difference.size();
        if (nSize_diff == 0)
            return 1;
            
        int nTmp = difference[0];
        int nResult = 1;
        for (int i = 1; i < nSize_diff; i++) {
            if ((nTmp > 0 && difference[i] < 0) || (nTmp < 0 && difference[i] > 0)) {
                ++nResult;
                nTmp = difference[i];
            }
        }
        return nResult + 1;
    }