[刷题防痴呆] 0376 - 摆动序列 (Wiggle Subs

题目地址

https://leetcode.com/problems/wiggle-subsequence/description/

题目描述

376. Wiggle Subsequence

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

思路

dp可解.

• 对于每一个位置. 有三种状态情况.
• 向上位置, nums[i] > nums[i-1].
• 向下位置, nums[i] < nums[i-1].
• 相等位置, nums[i] == nums[i-1].
• 因此我们可以使用两个数组来记录当前位置的最大的wiggle sequence length.
• 如果nums[i] > nums[i-1], 就是向上, 则之前必须为向下, up[i] = down[i-1] + 1, down数组不变.
• 如果nums[i] < nums[i-1], 就是向下, 则之前必须为向上, down[i] = up[i-1] + 1, up数组不变.
• 如果nums[i] == nums[i-1], 不会变. up[i] = up[i - 1], down[i] = down[i - 1], 原因是subsequence可以删除重复相等元素, 所以我们记录最大值.

关键点

• init up[0] = 1. down[0] = 1.
• 最后返回Math.max(down[nums.length-1],up[nums.length-1]).
• 可以优化空间为, 两个数组变为两个count来记录.

代码

• 语言支持：Java

class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }        
        int[] up = new int[nums.length];
        int[] down = new int[nums.length];
        
        up[0] = 1;
        down[0] = 1;
        
        for (int i = 1; i < nums.length; 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 Math.max(up[nums.length - 1], down[nums.length - 1]);
    }
}

// dp 优化
class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int n = nums.length;
        int up = 1;
        int down = 1;

        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) {
                up = down + 1;
            } else if (nums[i] < nums[i - 1]) {
                down = up + 1;
            } 
        }

        return Math.max(up, down);
    }
}