[LeetCode] 34. Search for a Range 搜索一个范围(Find First and Last Position of Element in Sorted Array)(代码
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原题目:Search for a Range, 现在题目改为: 34. Find First and Last Position of Element in Sorted Array
Given an array of integers nums 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].
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
给一个有序整数数组中,寻找相同目标值的起始和结束位置,限定了时间复杂度为O(logn)。
解法:二分法,典型的二分查找法的时间复杂度,先对原数组使用二分查找法,找出其中一个目标值的位置,然后向两边搜索找出起始和结束的位置。
Java:
public class Solution { public int[] searchRange(int[] A, int target) { int start = Solution.firstGreaterEqual(A, target); if (start == A.length || A[start] != target) { return new int[]{-1, -1}; } return new int[]{start, Solution.firstGreaterEqual(A, target + 1) - 1}; } //find the first number that is greater than or equal to target. //could return A.length if target is greater than A[A.length-1]. //actually this is the same as lower_bound in C++ STL. private static int firstGreaterEqual(int[] A, int target) { int low = 0, high = A.length; while (low < high) { int mid = low + ((high - low) >> 1); //low <= mid < high if (A[mid] < target) { low = mid + 1; } else { //should not be mid-1 when A[mid]==target. //could be mid even if A[mid]>target because mid<high. high = mid; } } return low; } }
Python:
class Solution(object): def searchRange(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ # Find the first idx where nums[idx] >= target left = self.binarySearch(lambda x, y: x >= y, nums, target) if left >= len(nums) or nums[left] != target: return [-1, -1] # Find the first idx where nums[idx] > target right = self.binarySearch(lambda x, y: x > y, nums, target) return [left, right - 1] def binarySearch(self, compare, nums, target): left, right = 0, len(nums) while left < right: mid = left + (right - left) / 2 if compare(nums[mid], target): right = mid else: left = mid + 1 return left def binarySearch2(self, compare, nums, target): left, right = 0, len(nums) - 1 while left <= right: mid = left + (right - left) / 2 if compare(nums[mid], target): right = mid - 1 else: left = mid + 1 return left def binarySearch3(self, compare, nums, target): left, right = -1, len(nums) while left + 1 < right: mid = left + (right - left) / 2 if compare(nums[mid], target): right = mid else: left = mid return left if left != -1 and compare(nums[left], target) else right
C++:
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { const auto start = lower_bound(nums.cbegin(), nums.cend(), target); const auto end = upper_bound(nums.cbegin(), nums.cend(), target); if (start != nums.cend() && *start == target) { return {start - nums.cbegin(), end - nums.cbegin() - 1}; } return {-1, -1}; } }; class Solution2 { public: vector<int> searchRange(vector<int> &nums, int target) { const int begin = lower_bound(nums, target); const int end = upper_bound(nums, target); if (begin < nums.size() && nums[begin] == target) { return {begin, end - 1}; } return {-1, -1}; } private: int lower_bound(vector<int> &nums, int target) { int left = 0; int right = nums.size(); // Find min left s.t. A[left] >= target. while (left < right) { const auto mid = left + (right - left) / 2; if (nums[mid] >= target) { right = mid; } else { left = mid + 1; } } return left; } int upper_bound(vector<int> &nums, int target) { int left = 0; int right = nums.size(); // Find min left s.t. A[left] > target. while (left < right) { const auto mid = left + (right - left) / 2; if (nums[mid] > target) { right = mid; } else { left = mid + 1; } } return left; } };
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