334. Increasing Triplet Subsequence

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.

遍历更新最小的数和次小的数，如果某个数大于这两个数，则有三个数递增，返回true.

class Solution {public:bool increasingTriplet(vector<int>& nums) {if (nums.size() < 3) return false;int firstMin = INT_MAX;int secondMin = INT_MAX;for (int i = 0; i < nums.size(); i++){if (nums[i] <= firstMin){//<=firstMin = nums[i];}else if (nums[i] <= secondMin){secondMin = nums[i];}else{return true;}}return false;}};