Search for a Range

标签：二分查找

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].

二分查找

class Solution {
public:
    int begin = -1, end = -1;
    vector<int> searchRange(int A[], int n, int target) {
       vector<int>ans;
       find(A,0,n-1,target);
       ans.push_back(begin);
       ans.push_back(end);
       return ans;
    }
    void find(int A[], int l, int r, int target){
      if(l > r) return ;
      int mid = (l+r) >> 1;
      if(A[mid] == target){
        if(begin == -1 || begin > mid)
          begin = mid;
        end = max(mid, end);
        find(A,l,mid-1,target);
        find(A,mid+1,r,target);
      }
      else if(A[mid] < target)
        find(A,mid+1,r,target);
      else 
        find(A,l,mid-1,target);
    }
};

Search for a Range

标签：二分查找

原文地址：http://blog.csdn.net/sina012345/article/details/43251243