334. Increasing Triplet Subsequence

题目：

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.

代码：

bool increasingTriplet(vector<int>& nums) {

    if(nums.size()<3) return false;

    int min = nums[0]; //
    int maxer = INT_MAX;
    for(int i=1;i<nums.size();i++){ //设置三种状态
        if(nums[i]<=min) min=nums[i]; //状态一：小于最小值，则更新最小值
        else if(nums[i]<maxer) maxer=nums[i]; //状态二：大于最小值，且小于次大值，则更新次大值
        else if(nums[i]>maxer) return true; //状态三：大于次大值，则代表有三个数是递增关系
    }

    return false;
}