bluewelkin 2020-04-09
给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
你的算法时间复杂度必须是?O(log n) 级别。
如果数组中不存在目标值,返回?[-1, -1]。
示例 1:
输入: nums = [5,7,7,8,8,10], target = 8 输出: [3,4]
示例?2:
输入: nums = [5,7,7,8,8,10], target = 6 输出: [-1,-1]
本题同【剑指Offer】面试题53 - I. 在排序数组中查找数字 I
时间复杂度:O(logn)
空间复杂度:O(1)
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { if (nums.empty()) return {-1, -1}; int left = searchLeft(nums, target); int right = searchRight(nums, target); return {left, right}; } int searchLeft(vector<int> &nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { if (mid == 0 || (mid - 1 >= 0 && nums[mid - 1] != target)) { return mid; } right = mid - 1; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return -1; } int searchRight(vector<int> &nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { if (mid == nums.size() - 1 || (mid + 1 <= nums.size() - 1 && nums[mid + 1] != target)) { return mid; } left = mid + 1; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return -1; } };
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { if (nums.empty()) return {-1, -1}; int left = searchLeft(nums, target); int right = searchRight(nums, target); return {left, right}; } int searchLeft(vector<int> &nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { //一直向左找 while (mid - 1 >= 0 && nums[mid - 1] == target) { --mid; } return mid; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return -1; } int searchRight(vector<int> &nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { //一直向右找 while (mid + 1 <= nums.size() - 1 && nums[mid + 1] == target) { ++mid; } return mid; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return -1; } };
lower_bound:返回一个迭代器,指向键值 >= key的第一个元素
upper_bound:返回一个迭代器,指向键值 > key的第一个元素
时间复杂度:O(logn)
空间复杂度:O(1)
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { if (nums.empty()) return {-1, -1}; auto left = lower_bound(nums.begin(), nums.end(), target); auto right = upper_bound(nums.begin(), nums.end(), target); if (left == right) return {-1, -1}; return {left - nums.begin(), right - nums.begin() - 1}; } };