LeetCode 34.Search for a Range

题目：

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]

.
分析与解答：看到题目就在想，这不就是STL里面的equalrange吗。依然是二分查找，只是查找停止的条件更严格了一些。可以分两次查找。第一次找左边界，第二次找右边界。

class Solution {
public:
vector<int> searchRange(int A[], int n, int target) {
int left = 0, right = n;
while(left != right) {
int mid = left + (right - left) / 2;
if(A[mid] >= target)
right = mid;
else
left = mid + 1;
}
int left1 = left;
left = 0, right = n;
while(left != right) {
int mid = left + (right - left) / 2;
if(A[mid] > target)
right = mid;
else
left = mid + 1;
}
if(A[left1] != target)//没找到target
return vector<int>{-1,-1};
else
return vector<int>{left1,left - 1};
}
};