leetcode 34. 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].

即在排序数组求给出成员的上下界。

用二分查找分别求出对应成员的上下界，但要注意比起普通的二分查找这儿的条件有所变化。

具体为对mid前后成员值的判断，还有防止出界的mid是否为vector首尾迭代器的条件判断。

可以while可以递归实现。

代码

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target){vector<int>::iterator start = nums.begin();vector<int>::iterator end = nums.end();vector<int>::iterator mid;vector<int> res;while (start <= end){mid = start + (end - start) / 2;int mid_data = *mid;if (mid_data == target){if ((mid != nums.begin() && *(mid - 1) != target) || mid == nums.begin())break;elseend = mid - 1;}else if (mid_data > target)end = mid - 1;elsestart = mid + 1;}if (*mid == target)res.push_back(mid - nums.begin());else{res.push_back(-1);res.push_back(-1);return res;}start = nums.begin();end = nums.end();while (start <= end){mid = start + (end - start) / 2;int mid_data = *mid;if (mid_data == target){if ((mid != nums.end() - 1 && *(mid + 1) != target) || mid == nums.end() - 1)break;elsestart = mid + 1;}else if (mid_data < target)start = mid + 1;elseend = mid - 1;}res.push_back(mid - nums.begin());return res;}};