LeetCode解题报告（321)-- 376. Wiggle Subsequence

Problem

Given an integer array nums, return the length of the longest wiggle sequence.

A wiggle sequence is a sequence where 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.

A subsequence is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

Example 1:

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3

Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.

Example 2:

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3

Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

Example 3:

1 2	Input: nums = [1,2,3,4,5,6,7,8,9] Output: 2

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] <= 1000

Follow up: Could you solve this in O(n) time?

Analysis

题目给出一个数组，要求我们找出最长的摆动子序列。摆动的定义是两者之差在正负摆动。首先这类在数组中找子序列长度的题目一般就采用dp来做，但是这里的摆动怎么定义状态呢？实际上这个和股票买卖是有点类似的，一天买一天卖。我们分别定义两个方程：

up[i]表示第i个数字处于上摆时，所能构成的摆动序列的最大长度；
down[i]表示第i个数字出于下摆时，所能构成的摆动序列的最大长度。

这样一来，我们就可以根据前一个位置来推出后一个位置的值了。我们根据前后两个数字值的大小关系来分别处理：

nums[i] > nums[i - 1]，说明后一个位置大，所以是上摆，结果就是前一个位置处于下摆的值加上1。同时因为这个位置不能下摆，所以下摆的值和前一个位置保持一致；
同理nums[i] < nums[i - 1]，说明前一个位置大，所以是下摆，结果就是前一个位置处于上摆的值加上1。因为这个位置不能上摆，所以上摆的值和前一个位置上摆的值保持一直。
如果两者相等，则和前一个位置的值都保持一致。

Solution

无。

Code

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

Summary

这道题也算是比较简单的dp题目，这类上下摆动、买卖的题目，很明显的特征就是，不但依赖于前一个状态，还要判断不同情况的处理逻辑。解决这类问题使用两个动态转移方程即可。这道题目的分享到这里，感谢你的支持！