Array - 376. Wiggle Subsequence

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

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

思路：

摇摆数组的最大值这题是很经典的动态规划，状态转移方程有两个,保存的分别是当前位置之前上凸下凹两种极点的个数，这样来找出最大的摇摆数组长度。

代码：

java：

class Solution {

    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        
        int len = nums.length;
        int[] down = new int[len];  // num down 
        int[] up = new int[len];
        
        // 初始条件
        down[0] = 1;
        up[0] = 1;
        
        for (int i = 1; i < len; i++) {
            if (nums[i] < nums[i-1]) {
                down[i] = up[i-1] + 1;
                up[i] = up[i-1];
            } else if (nums[i] > nums[i-1]) {
                up[i] = down[i-1] + 1;
                down[i] = down[i-1];
            } else {
                // 两数相等
                up[i] = up[i-1];
                down[i] = down[i-1];
            }
        }
        
        return Math.max(up[len - 1], down[len - 1]);
    }
}