LeetCode 376. Wiggle Subsequence

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: 6The entire sequence is a wiggle sequence.Input: [1,17,5,10,13,15,10,5,16,8]Output: 7There 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

解题思路：动态规划。用length数组记录从开始到每个数组元素的最大 wiggle sequence 长度，例如数组[1,4,3,5]对应的length数组为[1,2,3,4];用positive数组记录每个元素与其在wiggle sequence 中的前一元素的大小关系，1表示比前一元素大，-1表示比前一元素小，0表示相等，例如数组[1,5,3,2,7]对应的positive数组为[0,1,-1,-1,1]。关键在于length[i]的更新，每次遍历nums[0]到nums[i-1]，当nums[i]能并入nums[j] (j=0,..,i-1)为最后元素的wiggle sequence 时，比较length[i]与length[j]+1的大小，若>则更新，否则不更新length[i]。最后返回length数组中的最大值。该做法的空间复杂度为O（n）,时间复杂度为O（n平方）。

代码如下：

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {        vector<int> length(nums.size(),1), positive(nums.size(),0);        int max = 1, temp = 0;        if(nums.size()==0) return 0;        for(vector<int>::size_type i = 0; i != nums.size(); ++i){            for(vector<int>::size_type j = 0; j != i; ++j){                temp = nums[i]-nums[j]>0?1:nums[i]-nums[j]==0?0:-1;                if(j == 0&&temp){                    if(length[j]+1 >= length[i]){                        length[i] = length[j] + 1;                        positive[i] = temp>0?1:-1;                    }                }                if(temp-positive[j]==2||temp-positive[j]==-2){                    if(length[j]+1 >= length[i]){                        length[i] = length[j] + 1;                        positive[i] = temp>0?1:-1;                    }                }            }            if(max<length[i]) max =length[i];         }        return max;    }};