LeetCode 376. Wiggle Subsequence（动态规划）

Wiggle Subsequence

Medium

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

题意

求数组中的最大交替子列长度

思路

动态规划，因为交替子列有两种情况：

1. 奇数元素比偶数元素大
2. 偶数元素比奇数元素大

因此用两个动态规划数组up[i], down[i]表示子问题[0:i]的最大交替子列长度
与『最大上升子列』问题不同，可以很容易证明，up[i]仅仅与up[i-1]或down[i-1]有关，down[i]仅仅与down[i-1]或up[i-1]有关，因此up和down数组可以压缩为两个整数

代码

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