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: 2Follow up: Can you do it in O(n) time?
解析：O(n)的复杂度，分析情况，从当前位置到下个位置如果是递增的则找到增加到最大的元素，再减小，减小的话也找到连续的最小的元素为当前减小的元素，并记录每次增加或者减小的方向。
代码：
class Solution { public: int wiggleMaxLength(vector<int>& nums) { if (nums.empty()) return 0; if (nums.size()==1) return 1; if (nums.size()==2) { if (nums[0]==nums[1]) return 1; return 2; } int ans=0; int cur; int flag=INT_MIN; int i; for ( i=1; i<nums.size(); i++) { if (nums[i]!=nums[0]) break; } if (i==(nums.size())) return 1; flag=nums[0]-nums[i]; cur=nums[i]; ans=2; i++; for (; i<nums.size(); i++) { if (flag<0)//上次递增 { if (nums[i]>=cur) { cur=nums[i]; } else { ans++; flag=1; cur=nums[i]; } } else { if (nums[i]<=cur) { cur=nums[i]; } else { ans++; flag=-1; cur=nums[i]; } } } return ans; } };
