题目链接：https://leetcode.com/problems/increasing-triplet-subsequence/?tab=Description
Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.
Formally the function should:
Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.
Examples: Given [1, 2, 3, 4, 5], return true.
Given [5, 4, 3, 2, 1], return false.
解题思路：
初始化c1,c2为INT_MAX，
（1）找第一个数a1时，如果a1<=c1，则c1=a1
（2）找第二个数a2时，如果c1<a2<=c2，则c2=a2
（3）找第三个数a3时，如果a3>c2 且a3>c1，则返回true
（4）找第二个数a2时，如果a2<=c1，则c1=a2
（5）找第三个数a3时，如果a3>c1且a3<=c2,则c2=a3
class Solution{ public: bool increasingTriplet(vector<int>& nums) { int c1=INT_MAX; int c2=INT_MAX; for(int i=0;i<nums.size();i++) { if(nums[i]<=c1) c1=nums[i]; else if(nums[i]<=c2) c2=nums[i]; else return true; } return false; } };
