[leetcode] 376. Wiggle Subsequence

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本文介绍了一种寻找最长摆动子序列的方法，包括两种解题思路：一种是通过判断拐点来确定序列长度，另一种是利用动态规划的思想解决。文章还提供了具体的代码实现。

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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

Follow up:
Can you do it in O(n) time?

这道题是找最长“摆动”序列的长度，题目难度为Medium。

首先我们需要考虑哪些元素在构成“摆动”序列时起到了关键作用，在序列向前延伸过程中，拐点的元素满足“摆动”条件，也就是说每遇到一个拐点就可以在“摆动”序列中加入该元素，而在加入了所有拐点元素之后其他元素并不能增长“摆动”序列的长度，所以题目就转化为找拐点元素的个数了。需要注意的是相等的连续元素，在拐点判断中这些元素可以视为一个元素，遇到相等的连续元素跳过即可。具体代码：

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        if(nums.empty()) return 0;
        int flag = -1, length = 1, prev = nums[0];
        
        for(int i=0; i<nums.size(); ++i) {
            if(nums[i] == prev) continue;
            if(flag != nums[i]>prev) {
                flag = nums[i]>prev;
                ++length;
            }
            prev = nums[i];
        }
        
        return length;
    }
};

看了下别人的代码，还可以通过动态规划的思路来解决问题。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，遍历数组，如果nums[i]>nums[i-1]，表明第i-1到第i个元素是上升的，因此up[i]只需在down[i-1]的基础上加1即可，而down[i]保持down[i-1]不变；如果nums[i]<nums[i-1]，表明第i-1到第i个元素是下降的，因此down[i]只需在up[i-1]的基础上加1即可，而up[i]保持up[i-1]不变；如果nums[i]==nums[i-1]，则up[i]保持up[i-1]，down[i]保持down[i-1]。比较最终以上升沿和下降沿结束的最长“摆动”序列长度即可获取最终结果。具体代码：

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