376 Wiggle Subsequence

最新推荐文章于 2020-06-30 21:00:14 发布

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最新推荐文章于 2020-06-30 21:00:14 发布

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分类专栏： LeetCode # Medium # DynamicProgramming # Solved # Greedy

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博客围绕求整数序列中最长摆动子序列长度展开。摆动序列要求相邻数差值正负交替。通过分析可知这是典型的动态规划问题，需考虑小项开头和大项开头两种数列情况，对任意元素向左寻找能连接到两种数列的最近元素，最后给出动态规划结果。

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1 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.

Example 1:

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

Example 2:

Input: [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:

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

2 尝试解

2.1 分析

给定一个整数向量，问不改变元素相对顺序的情况下，能构成的最大振荡数列的长度。

振荡数列随向量的扩张是连续变化的，所以这是一个典型的动态规划问题，最长子列可能是小→大→小→大，那么奇数项是小项，偶数项是大项；也可能是大→小→大→小，那么奇数项是大项，偶数项是小项。所以，对于任意元素nums[i]，向左寻找能连接到两种数列的最近一个元素。如对于元素15，如果放在小项开头的数列中，

如果作为偶数项加入，则要找到前一个比他小的元素，且该元素在数列中为奇数项，即元素5
如果作为奇数项加入，则要找到前一个比他大的元素，且该元素在数列中为偶数项，即元素17

只要找到就停止寻找，所以找到5就会停止，即15作为第4项加在小项开头数列中。大项开头的数列同理。

动态规划完成，结果如下所示：

1 17 5 10 13 15 10 5 16 8

小项开头 1 2 3 4 4 4 5 5 6 7

大项开头 1 1 2 3 3 3 4 4 5 6

2.2 代码

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        vector<int> first_big(nums.size(),1);
        vector<int> first_small(nums.size(),1);
        int result = nums.size()? 1 : 0;
        for(int i = 1; i < nums.size(); i++){
            for(int j = i - 1; j >= 0; j--){
                if((nums[j] > nums[i] && first_small[j] % 2 == 0)||(nums[j] < nums[i] && first_small[j] % 2 == 1)){
                    first_small[i] = first_small[j] + 1;
                    if(first_small[i] > result)
                        result = first_small[i];
                    break;
                }
            }
            for(int j = i - 1; j >= 0; j--){
                if((nums[j] > nums[i] && first_big[j]%2 == 1)|| (nums[j] < nums[i] && first_big[j] % 2 == 0)){
                    first_big[i] = first_big[j] + 1;
                    if(first_big[i] > result)
                        result = first_big[i];
                    break;
                }
            }
        }
        return result;
    }
};

3 标准解

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