代码 // Search in Rotated Sorted Array II// Time Complexity: O(n),Space Complexity: O(1)publicclassSolution{publicbooleansearch(int[]nums,inttarget){intfirst=0,last=nums.length;while(first!=last){finalintmid=first+(last-first)/2;if(nums[mid]==target)returntrue;if(nums[first]<nums[mid...
既然是由两个有序数组拼接而成,那么只需找到断点,然后进行两次二分查找就行. Time complexity: O(N) view code /** * --- * Copyright (c) 2016 crazyacking.All rights reserved. * --- * Author: crazyacking * Date : 2016-03-01-21.22 */ #include <queue> #include <cstdio> #include <s...
三:Search in Rotated Sorted ArrayII Follow up for "Search in Rotated Sorted Array": What ifduplicatesare allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 数组中如果有重复元素时,时间复杂度退化到o(n)。 如...
Given the arraynumsafterthe possible rotation and an integertarget, returnthe index oftargetif it is innums, or-1if it is not innums. You must write an algorithm withO(log n)runtime complexity. Example 1: Input:nums = [4,5,6,7,0,1,2], target = 0Output:4 Example 2: Input:nums...
Search in Rotated Sorted Array极为相似,而且题目要求我们的时间复杂度也是O(log n),根据时间复杂度的提醒我们可以想到二分法。 代码说明: 代码先使用经典二分法找到一个符合的元素,然后再建立指针左右寻找符合的边界,代码中已有注释,请参考注释。 测试代码:......
Follow up for "Search in Rotated Sorted Array": What if duplicates Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. class Solution { public: //1,1, 1,2,1,1,1,恰好左中右都是1,这时我们 ++左下标,--右下标 ...
Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array. 本题难度Medium。 【复杂度】 时间O(N) 空间 O(1) 【思路】 本题是Search in Rotated Sorted Array的变形。在Search in Rotated Sorted Array中可以通过nums[sta...
81 Search in Rotated Sorted Array II 详细解答 由于题目要求时间复杂度是O(log n), 很容易想到二分法。 但应该怎么构造二分? 举例: nums = [4,5,6,7,0,1,2], target = 0 step 1: l, r = 0, 6, mid = 3, 此时 nums[3] 先跟 nums[l] 比较 如果 nums[3] > nums[... ...
You are given a target value to search. If found in the array return its index, otherwise return -1.You may assume no duplicate exists in the array.Your algorithm's runtime complexity must be in the order of O(log n).Example 1:Input: nums = [4,5,6,7,0,1,2], target = 0 ...
Search in Rotated Sorted Array II Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to ...[leetcode]Search in Rotated Sorted Array 新博文地址:[leetcode]Search in Rotated Sorted Arra...