[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].
 
分析：二分查找
 1 class Solution
 2 {
 3 public:
 4   void Searching(vector<int> &ret, int A[], int s, int e, int target)
 5   {
 6     if(s > e)
 7     {
 8       ret.push_back(-1);
 9       ret.push_back(-1);
10       return;
11     }
12 
13     int mid = (s+e)/2;
14     if(A[mid] < target)
15       Searching(ret, A, mid+1, e, target);
16     else if(A[mid] > target)
17       Searching(ret, A, s, mid-1, target);
18     else
19     {
20       int i = mid - 1;
21       while(i >= s && A[i] == A[mid]) i--;
22       ret.push_back(i + 1);
23 
24       i = mid + 1;
25       while(i <= e && A[i] == A[mid]) i++;
26       ret.push_back(i - 1);
27     }
28   }
29   vector<int> searchRange(int A[], int n, int target)
30   {
31     vector<int> ret;
32     if(target < A[0] || target > A[n-1])
33     {
34       ret.push_back(-1);
35       ret.push_back(-1);
36     }
37     else
38       Searching(ret, A, 0, n-1, target);
39 
40     return ret;
41   }
42 };

 

posted @ 2015-03-17 15:30  imKirin  阅读(144)  评论(0)  编辑  收藏  举报