leetcode [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?

題目大意：

找出最長的擺動序列的長度。

解法：

使用兩個動態規劃數組up和down。

up[i]記錄以nums[i]結尾的子序列的長度（該子序列的最後兩個元素是遞增的）。

down[i]記錄以nums[i]結尾的子序列的長度（該子序列的最後兩個元素是遞減的）。

class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums.length == 0) return 0;
        int[] up=new int[nums.length];
        int[] down=new int[nums.length];
        up[0]=1;
        down[0]=1;
        for (int i=1;i<nums.length;i++){
            if (nums[i]>nums[i-1]){
                up[i]=down[i-1]+1;
                down[i]=down[i-1];
            }else if (nums[i]<nums[i-1]){
                down[i]=up[i-1]+1;
                up[i]=up[i-1];
            }else{
                down[i]=down[i-1];
                up[i]=up[i-1];
            }
        }

        return Math.max(down[nums.length-1],up[nums.length-1]);
    }
}