334. Increasing Triplet Subsequence My Submissions Question--Avota

问题描述:

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.

题意:

给定一无序数组,判断其中是否存在长度为3的递增子序列。

解题思路:

此问题可转化为先找长度为2的递增序列(记为first, second)然后再判断后面是否有比second更大的数。此时我们只需维护first, second,但可能出现如3,4,1这种情况,这时我们就需要另一个数temp表示位于递增序列后面但比first更小的数。之后对于数组中的每一个数nums[i],只需根据其在temp,first,second三者之间的位置来判断结果并维护这三个数,其中包括[负无穷, temp],(temp, first],(first, second],(second, 正无穷)四个区间

示例代码:

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        int len = nums.size();
        if(len < 3){
            return false;
        }
        int first = nums[0], second = INT_MAX, temp = nums[0];
        for(int i = 1; i < len; i++){
            if(nums[i] > second){
                return true;
            }
            if(nums[i] > first){
                second = nums[i];
            }
            else if(nums[i] <= temp){
                temp = nums[i];
            }
            else{
                first = temp;
                second = nums[i];
            }
        }
        return false;
    }
};
posted @ 2016-03-06 22:04  Avota  阅读(156)  评论(0编辑  收藏  举报