[Leetcode] #334 Increasing Triplet Subsequence

最新推荐文章于 2018-09-29 20:09:52 发布

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最新推荐文章于 2018-09-29 20:09:52 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/qq_34264472/article/details/71403887

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本文介绍了一种在未排序数组中查找长度为3的递增子序列的方法，并提供两种解决方案。一种是利用最长递增子序列(LIS)思想但会超时，另一种是在O(n)时间内和O(1)空间复杂度下完成任务。

Discription:

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:

bool increasingTriplet(vector<int>& nums) { //借鉴LIS，看长度是否大于3，超时
	int len = nums.size();
	if (len < 3)
		return false;
	vector<int> dp(len, 1);
	for (int i = 1; i < len; i++){
		for (int j = i - 1; j >= 0; j--){
			if (nums[i]>nums[j]){
				dp[i] = max(dp[i], dp[j] + 1);
			}
		}
		if (dp[i] >= 3)
			return true;
	}
	return false;
}

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

GitHub-Leetcode: https://github.com/wenwu313/LeetCode