Search for a Range

最新推荐文章于 2022-03-08 23:18:06 发布

水晶莲

最新推荐文章于 2022-03-08 23:18:06 发布

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分类专栏： Leetcode 文章标签： Leetcode

本文链接：https://blog.youkuaiyun.com/a1231988/article/details/38980073

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博客围绕在排序数组中查找给定目标值的起始和结束位置展开，要求算法时间复杂度为O(log n)，若目标值不在数组中则返回特定结果，并给出了示例。

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Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

 vector<int> searchRange(int A[], int n, int target) {
	int low =0,high =n-1,index,start =-1,end =1;
	vector<int> vec;
	int mid;
	while(low<=high)
	{
		mid =(low+high)/2;
		if(A[mid]>target)
		{
			high =mid-1;
		}
		else if(A[mid]<target)
		{
			low =mid +1;
		}
		else
		{
			index =mid;
			break;
		}
	}
	if(low<=high)
	{
		start =end =mid;
		while(start>=0&&A[--start] ==target);
		if(start<0)
		{
			vec.push_back(0);
		}
		else
		{
			vec.push_back(start+1);
		}
		while(end<=n-1&&A[++end] ==target);
		if(end>n-1)
		{
			vec.push_back(n-1);
		}
		else
		{
			vec.push_back(end-1);  
		}
		return vec;
	}
	else
	{
		vec.push_back(-1);
		vec.push_back(-1);
		return vec;
	}
}