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.

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

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

一刷
题解：
采用discuss中的思路：

Step 1: First we check our requirement is to get small number. As 1<2 so the series will be
 2,1
Step 2: Now we need big number that is  greater than 1. As 4>1 so series  will be
2,1,4
Step 3: Now we need small number. But 5>4 so 4 will be replaced by 5. So the series will become
2,1,5
Step 4:  We need small number. But 6>5. Series will be
2,1,6
Step 5: Require small number. 3<6. Series will be
2,1,6,3
Step 6: Require big number. 3=3. No change in series
2,1,6,3
Step 7: Require big number. 4>3. Series will become
2,1,6,3,4
Step 8:  Require small number. 8>4. 8 will  replace 4 and series will become
2,1,6,3,8
Step 9: Require small number. 4<8. So final series will  be
2,1,6,3,8,4

public class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums.length == 0 || nums.length == 1) {
            return nums.length;
        }
        int k = 0;
        while (k < nums.length - 1 && nums[k] == nums[k + 1]) {  //Skips all the same numbers from series beginning eg 5, 5, 5, 1
            k++;
        }
        if (k == nums.length - 1) {
            return 1;
        }
        int result = 2;     // This will track the result of result array
        boolean smallReq = nums[k] < nums[k + 1];       //To check series starting pattern
        for (int i = k + 1; i < nums.length - 1; i++) {
            if (smallReq && nums[i + 1] < nums[i]) {
                nums[result] = nums[i + 1];
                result++;
                smallReq = !smallReq;    //Toggle the requirement from small to big number
            } else {
                if (!smallReq && nums[i + 1] > nums[i]) {
                    nums[result] = nums[i + 1];
                    result++;
                    smallReq = !smallReq;    //Toggle the requirement from big to small number
                }
            }
        }
        return result;
    }
}