Search for a Range

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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].

思路：使用二分查找即可。调用标准库函数equal_range函数：

 1 class Solution {
 2 public:
 3     vector<int> searchRange( int A[], int n, int target ) {
 4         vector<int> ret( 2, -1 );
 5         pair<int*,int*> range = equal_range( A, A+n, target );
 6         if( range.first != range.second ) {
 7             ret[0] = range.first-A;
 8             ret[1] = range.second-A-1;
 9         }
10         return ret;
11     }
12 };

Search for a Range

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原文地址：http://www.cnblogs.com/moderate-fish/p/3954405.html