leetcode376. Wiggle Subsequence

CoffeX 发布于2019-08-16 17:10 / 3321人阅读

摘要：题目要求扭动序列是指数组中的相邻两个元素的差保证严格的正负交替，如数组中相邻两个元素的差为，满足扭动序列的要求。现在要求从一个数组中，找到长度最长的扭动子序列，并返回其长度。即前一个元素和当前元素构成下降序列，因此代码如下

题目要求

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

扭动序列是指数组中的相邻两个元素的差保证严格的正负交替，如[1,7,4,9,2,5]数组中相邻两个元素的差为6,-3,5,-7,3，满足扭动序列的要求。现在要求从一个数组中，找到长度最长的扭动子序列，并返回其长度。

思路和代码

这是一个可以通过动态规划来解决的问题。动态规划的特点就是，加入我知道第i个元素的结果，那么第i+1个元素的结果可以由其推到出来。这里假设我们知道，以第i个元素为止的最长子序列长度，包括上升序列up和下降序列down，则第i+1个元素的可能情况如下：

nums[i+1]>nums[i]: 即前一个元素和当前元素构成上升序列，因此up[i+1]=down[i]+1, down[i+1]=down[i]，这是指以第i个元素为结尾的上升序列应当基于第i-1个元素为结尾的下降序列，而以第i个元素为结尾的下降序列，等同于基于第i-1个元素为结尾的下降序列。

nums[i+1]>nums[i]: 即前一个元素和当前元素构成下降序列，因此down[i+1]=up[i]+1, up[i+1]=up[i]

nums[i+1]=nums[i]: down[i+1]=down[i], up[i+1]=up[i]

代码如下：

    public int wiggleMaxLength(int[] nums) {
        if( 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[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 Math.max(up[nums.length-1], down[nums.length-1]);
    }

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