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].

边界条件很恶心。

主要思想是：

1.左边界的前一个数要小于target，右边界的后一个数要大于target

2.如果没有边界左（右）的数，则可以直接返回

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> re;
        re.push_back(-1);
        re.push_back(-1);
        if(n < 1)
        {

            return re;
        }
        re[0]=findleft(A,n,target);
        re[1]=findright(A,n,target);
        return re;
        

    }
    
    int findleft(int A[],int n,int target)
    {
        int left = 0; 
        int right = n-1;
        if(A[0]==target)return 0;
        while(left <= right)
        {
            int mid = left + (right - left)/2;
            if(A[mid] == target &&( mid-1 >=0 &&A[mid-1] < target|| mid ==0))return mid;
            if(A[mid] >= target )//&& mid-1 >=0 &&A[mid-1] == target)
            {
                right = mid -1;
            }
            else if(A[mid] < target)
            {
                left = mid+1;
            }
        }
        return -1;
    }
    int findright(int A[],int n,int target)
    {
        int left = 0; 
        int right = n-1;
        if(A[n-1]==target)return n-1;
        while(left <= right)
        {
            int mid = left + (right - left)/2;
            if(A[mid] == target && (mid+1 <n &&A[mid+1] > target || mid+1 == n))return mid;
            if(A[mid] > target )
            {
                right = mid -1;
            }
            else if(A[mid] <= target )//&& mid+1<n && A[mid+1]==target )
            {
                left = mid+1;
            }
        }
        return -1;
    }
};