34. Search for a Range

Given an array of integers sorted in ascending order, 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].

题目规定必须在O（logn）的时间复杂度完成这道题，最简单的方法显然是二分搜索。我的思路是，先用二分搜索找到一个目标的下标，然后以该下标为中心，分别向前向后遍历，从而求出范围。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ans;        int index = binarySearch(0, nums.size() - 1, nums, target);        if (index == -1) {            ans.push_back(-1);            ans.push_back(-1);            return ans;        }        for (int i = index; i >= 0; i--) {            if (nums[i] != target) {                ans.push_back(i + 1);                break;            }        }        if (nums[0] == target) ans.push_back(0);        for (int j = index; j < nums.size(); j++) {            if (nums[j] != target) {                ans.push_back(j - 1);                break;            }        }        if (nums[nums.size() - 1] == target) ans.push_back(nums.size() - 1);        return ans;    }    int binarySearch(int begin, int end, vector<int> nums, int target) {        if (end < begin || nums.empty()) return -1;        int mid = begin + (end - begin) / 2;        if (nums[mid] == target) return mid;        if (target < nums[mid]) return binarySearch(begin, mid - 1, nums, target);        if (target > nums[mid]) return binarySearch(mid + 1, end, nums, target);    }};