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);elsereturn binary(array, low, mid - 1, key);}};