376. Wiggle Subsequence

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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:

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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

解法1：两次dp

两次dp的思想很巧妙，用up和down两个数组记录到当前位置，最后相邻的两个是up的最长wiggle sequence或者是最后相邻两个是down的最长wiggle sequence。
那么只要交替更新一下每一个dp array就可以了。
似乎这和minmax类的题目很类似， 比如Predict the winner，也是用两个相互影响的dp数组来完成。
C++

1

Java

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