标签：c++   leetcode   binary search   array   

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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].

//vs2012测试代码
//Binary Search
//we must know how to process the 3 cases below:
//1.how to find the right most target
//2.how to find the left most target
//3.how to find the insert position
#include<iostream>
#include<vector>

using namespace std;

#define N 6

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ans(2);
		if(n==0)
			return vector<int>( 2,-1 );
		int left = 0 , right = n-1;
		while( left<=right )
		{
			int mid = left + (right-left)/2;
			if( A[mid] >= target)  //find the left most target
				right = mid-1;
			else if (A[mid] < target)
				left = mid+1;
		}
		int target_left = left;

		left=0 , right=n-1;
		while( left<=right )
		{
			int mid = left + (right-left)/2;
			if(A[mid] <= target) //find the right most target
				left = mid +1;
			else if ( A[mid] > target)
				right = mid -1;
		}
		int target_right = right;

		if( A[target_left]!=target || A[target_right]!=target)
			return vector<int>( 2,-1 );
		ans[0] = target_left;
		ans[1] = target_right;
		return ans;
    }
};

int main()
{
	int a,target;
	int A[N];
	vector<int> ans(2);
	for(int i=0; i<N; i++)
	{
		cin>>a;
		A[i]=a;
	}
	cin>>target;
	Solution lin;
	ans = lin.searchRange( A,N,target);
	for(int i=0; i<2; i++)
		cout<<ans[i];
	cout<<endl;
}

//方法一：自测Accepted
//Binary Search
//we must know how to process the 3 cases below:
//1.how to find the right most target
//2.how to find the left most target
//3.how to find the insert position
class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ans(2);
		if(n==0)
			return vector<int>( 2,-1 );
		int left = 0 , right = n-1;
		while( left<=right )
		{
			int mid = left + (right-left)/2;
			if( A[mid] >= target)  //find the left most target
				right = mid-1;
			else if (A[mid] < target)
				left = mid+1;
		}
		int target_left = left;

		left=0 , right=n-1;
		while( left<=right )
		{
			int mid = left + (right-left)/2;
			if(A[mid] <= target) //find the right most target
				left = mid +1;
			else if ( A[mid] > target)
				right = mid -1;
		}
		int target_right = right;

		if( A[target_left]!=target || A[target_right]!=target)
			return vector<int>( 2,-1 );
		ans[0] = target_left;
		ans[1] = target_right;
		return ans;
    }
};

leetcode_34_Search for a Range

标签：c++   leetcode   binary search   array   

原文地址：http://blog.csdn.net/keyyuanxin/article/details/43635879