[Leetcode] #34 Search for a Range

Discription:

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].

Solution:

//中间找到一点 向两边扩vector<int> searchRange1(vector<int>& nums, int target) {vector<int> ans = { -1, -1 };if (nums.empty()) return ans;int left = 0, right = nums.size() - 1;while (left <= right){int mid = (left + right) >> 1;if (nums[mid] == target){ans[0] = mid;ans[1] = mid;break;}if (nums[mid] < target)left = mid + 1;elseright = mid - 1;}if (left>right)return ans;int i;for (i = ans[1]; i < nums.size();){if (nums[i] == target)i++;elsebreak;}ans[1] = i - 1;for (i = ans[0]; i >= 0;){if (nums[i] == target)i--;elsebreak;}ans[0] = i+1;return ans;}

//两次二分搜索 分别找最左点和最右点vector<int> searchRange(vector<int>& nums, int target) {vector<int> ans(2, -1);if (nums.empty()) return ans;int left = 0, right = nums.size() - 1;while (left < right){int mid = (left + right) >> 1;if (nums[mid] < target) left = mid + 1;else right = mid;}if (nums[left] != target) return ans;else ans[0] = left;right = nums.size() - 1;while (left < right){int mid = ((left + right) >> 1) + 1;if (nums[mid] > target) right = mid - 1;else left = mid;}ans[1] = right;return ans;}

附：Leetcode源代码见我的GitHub