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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].
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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