【leetcode】Search for a Range

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].

 

 

 

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
       
 
        vector<int>res(2);          
 
        res[0]=bs(A,n,target-1)+1;
        res[1]=bs(A,n,target);        
  
        if(res[1]==-1||A[res[1]]!=target)
        {
            res[0]=-1;
            res[1]=-1;
        }          
        return res;
    }
   
    //通过这个二分查找，如果有多个target的话可以找到最靠右边的元素
    //同时也得注意，如果没有target则找到的是比target小的最大的最靠右元素
     int bs(int A[],int n,int target)
    {
        int left=0;
        int right=n-1;
        int mid=(left+right)/2;
        int ret=-1;
       
        while(left<=right)
        {
            if(A[mid]>target)
            {
                right=mid-1;
            }
            else
            {
                //只要是当前元素小于等于target，left就会右移，因此找到最靠右的元素
                ret=mid;
                left=mid+1;
            }
            mid=(left+right)/2;
        }          
        return ret;
    }
   
};