leetcode 376.Wiggle Subsequence （greedy O(n))_wiggle subsequence greedy-优快云博客

本文介绍了一种求解最长摇摆子序列的算法，通过分析数组中元素的升降趋势，利用贪心策略或动态规划思想，记录上升和下降序列的长度，最终找到最长的摇摆子序列长度。

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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.

给出一个数组，让判断最长的摇摆子序列长度，即元素保持一升一降

思路：
greedy 或者dp
感觉greedy也可以说是简化版的dp

记录下两个方向up和down，用变量表示，（dp的话就是保存下所有的up，down值，数组长度）

初始状态up，down都是1
比如[1,2,3,4,2]
刚开始到2的时候上升，所以up=down + 1 = 2
然后3和4的时候一路上升，因为只是上升，趋势没有变，down仍然停留在1，up由down决定，也不变
后面到最后一个2的时候，下降，下降次数由上一个上升次数决定，down=up+1
（因为是摇摆序列，没有上升就没有下降，没有下降就没有上升）

注意当数组长度是0的时候，因为up和down初始值都是1，所以取min(n, max(up, down))

    public int wiggleMaxLength(int[] nums) {
       if (nums == null) {
           return 0;
       } 
        
        int up = 1;
        int down = 1;
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > nums[i - 1]) {
                up = down + 1;
            } else if (nums[i] < nums[i - 1]) {
                down = up + 1;
            }
        }
        
        return Math.min(nums.length, Math.max(up, down));
    }