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].

#include<iostream>
#include<vector>
using namespace std;

vector<int> searchRange(int A[], int n, int target) {
	int first = 0;
	int last  = n - 1;
	vector<int>result(2, -1);
	while (first<=last)
	{
		int mid = (first + last) / 2;
		if (A[mid] == target)
		{
			result[0] = mid;
			result[1] = mid;
			while (result[0]-1 >= first&&A[result[0]-1] == target)//当一位是重复位时才对范围跟新
				--result[0];	
			while (result[1]+1 <= last&&A[result[1]+1] == target)
				++result[1];
			return result;
		}
		else if (A[mid] < target)
			first = mid + 1;
		else
			last = mid - 1;
	}
	return result;
}

Search for a Range

原文：http://blog.csdn.net/li_chihang/article/details/44428087