leetcode 376. Wiggle Subsequence

最新推荐文章于 2021-07-22 15:40:24 发布

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本文介绍了一种寻找最长摆动子序列的有效算法。摆动序列是指序列中相邻元素之差严格交替正负的序列。文章详细解释了如何通过动态规划的方法来解决这一问题，并给出了具体的代码实现。

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?

这种有状态的题已经出现好多次了，大多数都可以设定状态转移的过程，每种状态设一个数组。
这里就是出于up 和 down 和equal的情况
解法如下：

public class Solution {
    public int wiggleMaxLength(int[] nums) {
        if(nums==null||nums.length==0) return 0;
        int len = nums.length;
        int[] up = new int[len];
        int[] down = new int[len];

        up[0] = 1;
        down[0] = 1;

        for(int i=1; i < len; 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[len-1], down[len-1]);
    }
}