Increasing Triplet Subsequence

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.

解析：

想了一会才想出来，用两个值表示连续三个数的第一个，第二个的值，下一个数在第一个与第二个之间时，更新第二个数为当前值，当前数小于第一个数时，更新第一个数为当前数，当前数大于第二数时找到结果。

代码：

class Solution {public:    bool increasingTriplet(vector<int>& nums) {        if (nums.size()<3)        return false;        int first,second;        second=INT_MAX;        int cnt=1;        first=nums[0];        for (int i=1; i<nums.size(); i++)        {           if (nums[i]>second)           return true;           if (nums[i]>first&&nums[i]<second)           {               second=nums[i];           }           else if (nums[i]<first)           {               first=nums[i];           }        }        return false;    }};