[leetcode] 334. Increasing Triplet Subsequence

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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.

Solution:

这里只需求出BIS3即可。

1、找到最小的数min1, 其位置为k1.

2、找到第二小的数min2, 其位置为k2 > k1.

3、如果k3（k3>k2）位置的数n>min2, 找到递增三元组(min1, min2, n)

　　如果n<=min1, 更新min1=n和k1 = k3, 如果n<=min2, 更新k2和min2.

 1 bool increasingTriplet(vector<int>& nums) 
 2     {
 3         int min1 = INT_MAX, min2 = INT_MAX;
 4         
 5         for (int i = 0; i < nums.size(); i++)
 6         {
 7             if (nums[i] <= min1)
 8                 min1 = nums[i];
 9             else if (nums[i] <= min2)
10                 min2 = nums[i];
11             else
12                 return true;
13         }
14         
15         return false;
16     }

[leetcode] 334. Increasing Triplet Subsequence

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原文地址：http://www.cnblogs.com/ym65536/p/5636523.html