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].

思路：二分查找法，找到目标值的范围，时间复杂度在O(logn)。先找出左边第一个等于目标值的位置，再找出右边最后一个等于目标值的位置，然后返回这个范围。

class Solution {

public:

    int findleft(int A[],int n,int target)

    {

        if(n==0)

            return -1;

        int left=0;

        int right=n-1;

        while(left<right)

        {

            int mid=(left+right)>>1;

            if(A[mid]>=target)

                right=mid;

            else

                left=mid+1;

        }

        if(target==A[left])

            return left;

        else

            return -1;

    }

    int findright(int A[],int n,int target)

    {

        if(n==0)

            return -1;

        int left=0;

        int right=n-1;

        while(left<=right)

        {

            int mid=(left+right)>>1;

            if(A[mid]>target)

                right=mid-1;

            else

                left=mid+1;

        }

        if(A[right]==target)

            return right;

        else 

            return -1;

    }

    vector<int> searchRange(int A[], int n, int target) {

        vector<int> result(2);

        result.clear();

        for(int i=0;i<2;i++)

            result.push_back(-1);

        int left=findleft(A,n,target);

        int right=findright(A,n,target);

        result[0]=left;

        result[1]=right;

        return result;

    }

};