220. Contains Duplicate III

最新推荐文章于 2019-05-01 19:22:10 发布

转载最新推荐文章于 2019-05-01 19:22:10 发布 · 242 阅读

本文介绍了一种算法，用于判断一个整数数组中是否存在两个不同的元素，这两个元素的索引之差不超过k，且数值之差不超过t。通过使用滑动窗口结合有序集合的方法，可以在O(n log n)的时间复杂度内解决此问题。

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k.

 bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t)
 {
    		if (!k || t<0 || nums.size()<2)
        		return false;
        	set<int>record;                   
        	auto nLen = nums.size();
        	for (int i = 0; i < nLen;++i)
        	{
        		if (i>k)
        			record.erase(nums[i - k - 1]);         
        		set<int>::iterator lower = record.lower_bound(nums[i] - t);
        		if (lower != record.end() && abs(nums[i] - *lower) <= t)
        			return true;
        
        		record.insert(nums[i]);
        	}
        	return false;
 }