leetcode376. Wiggle Subsequence

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题目要求

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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.

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?

扭动序列是指数组中的相邻两个元素的差保证严格的正负交替，如[1,7,4,9,2,5]数组中相邻两个元素的差为6,-3,5,-7,3，满足扭动序列的要求。现在要求从一个数组中，找到长度最长的扭动子序列，并返回其长度。

思路和代码

这是一个可以通过动态规划来解决的问题。动态规划的特点就是，加入我知道第i个元素的结果，那么第i+1个元素的结果可以由其推到出来。这里假设我们知道，以第i个元素为止的最长子序列长度，包括上升序列up和下降序列down，则第i+1个元素的可能情况如下：

  • nums[i+1]>nums[i]: 即前一个元素和当前元素构成上升序列，因此up[i+1]=down[i]+1, down[i+1]=down[i]，这是指以第i个元素为结尾的上升序列应当基于第i-1个元素为结尾的下降序列，而以第i个元素为结尾的下降序列，等同于基于第i-1个元素为结尾的下降序列。
  • nums[i+1]>nums[i]: 即前一个元素和当前元素构成下降序列，因此down[i+1]=up[i]+1, up[i+1]=up[i]
  • nums[i+1]=nums[i]: down[i+1]=down[i], up[i+1]=up[i]

代码如下：

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