leetcode 34:

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]

.
思路：

这题比较简单，可以先用二分法找到target的位置，然后向两边展开，找到相等元素的两个边界即可。

时间复杂度：O(lgn)

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int size = nums.size();
vector<int> result(2,-1);
if (size == 0) return result;
int index = binary(nums, 0, size - 1, target);
if (index == -1) return result;
int low = index, high = index;
while (low >=0 && nums[low] == target)
low--;
while (high < size && nums[high] == target)
high++;
result[0] = low+1;
result[1] = high-1;
return result;

}
int binary(vector<int>&array, int low, int high, int key)
{
if (low > high)
return -1;
int mid = (low + high) / 2;

if (array[mid] == key)
return mid;
else if (array[mid] < key)
return binary(array, mid + 1, high, key);
else
return binary(array, low, mid - 1, key);
}

};