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.

这题其实很简单，始终保持现有检索序列里的最小的两个符合递增要求的数就行。一旦发现比他俩都大的就返回true。

代码：

bool increasingTriplet(vector<int>& nums) {    int len = nums.size();    int min1 = INT_MAX;    int min2 = INT_MAX;    for (int i = 0; i < len; i++) {        if (nums[i] <= min1) {            min1 = nums[i];        }        else if (nums[i] <= min2) {            min2 = nums[i];        }        else return true;    }    return false;}