### 题目

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:

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


• Can you do it in O(n) time?

### 遍历数组，跳过单调递增或单调递减的元素即可

[1, 17, 5, 10, 13, 15, 10, 5, 16, 8]
16  -12 5  3   2   -5  -5 11  -8
|+++|--|+++|+++|++|----|--|+++|--|

1
1   -> 17               // 改变单调性就增加列表长度
1   -> 17  -> 5         // ...
1   -> 17  -> 5  -> 10  // ...
1   -> 17  -> 5  -> 13  // 出现连加，或连减，不增加链条长度，只更新最大最小值即可
因为13 > 10，所以用较大的13替换较小的10，不影响接下来的单调性


#### 迭代版

class Solution {
public int wiggleMaxLength(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int count = 1, sign = 0;
for (int i = 1; i < nums.length; i++) {
int diff = nums[i] - nums[i-1];
if ((sign >= 0 && diff < 0) || (sign <= 0 && diff > 0)) {
count++;
sign = diff;
}
}
return count;
}
}


#### 递归动态规划版

class Solution {
public int wiggleMaxLength(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
localNums = nums;
return Math.abs(dp(0)) + 1;
}
private int[] localNums;
private int dp(int index) {
if (index == localNums.length - 1) {
return 0;
}
int diff = localNums[index] - localNums[index + 1];
int sub = dp(index + 1);
int res = sub;
if (sub <= 0 && diff > 0) {
res = - (sub - 1);
} else if (sub >= 0 && diff < 0) {
res = - (sub + 1);
}
return res;
}
}


#### 结果 