334. Increasing Triplet Subsequence

标签：ret   exit   lock   orm   form   sorted   dex   思路   bsp   

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:c1表示包含当前点的情况下最小的待选index，c2表示包含当前点的情况下中间值的待选index

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
       int c1=INT_MAX,c2=INT_MAX;
       for(int num: nums){
           if(num<=c1)c1=num;
           else if(num<=c2)c2=num;
           else return true;
       }
       return false;
    }
};

334. Increasing Triplet Subsequence

标签：ret   exit   lock   orm   form   sorted   dex   思路   bsp   

原文地址：http://www.cnblogs.com/tsunami-lj/p/6536982.html