Leetcode--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].

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 Array Binary Search

Have you met this question in a real interview?

思路：返回的是一個數組，[左界，右界].下面就是尋找左界和右界

左界：在A[mid]==target時，如果mid=0或A[mid-1]!=target時，就觸碰到了邊界

右界：在A[mid]==target時，如果mid=n-1或A[mid+1]!=target時，觸碰邊界返回

class Solution {
public:
    int searchleft(int A[],int n,int target)
    {
        int l=0;
        int r=n-1;
        while(l<=r)
        {
            int mid=l+(r-l)/2;
            if(A[mid]==target)
            {
                if(mid==0)
                    return mid;
                
                if(A[mid-1]==target)
                    r=mid-1;
                else
                    return mid;
            }
            else if(A[mid]<target)
                l=mid+1;
            else 
                r=mid-1;
        }
        return -1;
    }
    int searchright(int A[],int n,int target)
    {
        int l=0;
        int r=n-1;
        while(l<=r)
        {
            int mid=l+(r-l)/2;
            if(A[mid]==target)
            {
                if(mid==n-1)
                    return mid;
                if(A[mid+1]==target)
                    l=mid+1;
                else
                    return mid;
            }
            else if(A[mid]>target)
                r=mid-1;
            else
                l=mid+1;
        }
        return -1;
    }
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> res;
        int l=0;
        int r=n-1;
        int start=searchleft(A,n,target);
        int end=searchright(A,n,target);
        res.push_back(start);
        res.push_back(end);
        return res;
    }
};