LeetCode之Search in Rotated Sorted Array II 1题目分析 本题是上面的升级版,即搜索含有重复字符的循环数组 2.解题思路 先看version1的题目吧,不含重复字符的循环数组搜索。 思路1. 暴力for循环 AC 思路2. 既然Rotate了,找到突变位置,将所有下标进行平移,再二分查找。 思路3. 其实可以不用平移的,我们看图就...
returnfalse; boolresult=searchD(nums, target, 0, size-1); returnresult; } };
题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ 题目: 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 determine if a given target is in the array. ...
本题是Search in Rotated Sorted Array的变形。在Search in Rotated Sorted Array中可以通过nums[start]<=nums[mid]或nums[mid]<=nums[end]判断两边是否为有序。但是在这里出现一个问题:比如左半边,如果nums[start]==nums[mid]它是有序吗?对于: [1...
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search. If found …
This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates. Would this affect the run-time complexity? How and why? 描述 假设按升序排序的数组在事先未知的某个枢轴处旋转。 (例如, [0,0,1,2,2,5,6] 可能变成 [2,5,6,0,0,1,2]). ...
Search in Rotated Sorted Array II 2. Solution class Solution{public:boolsearch(vector<int>&nums,inttarget){intsize=nums.size();intleft=0;intright=size-1;while(left<=right){intmid=(left+right)/2;if(nums[mid]==target){returntrue;}if(nums[left]==nums[mid]){left++;continue;}if(nums...
LeetCode-33-搜索旋转排序数组(Search in Rotated Sorted Array) 33. 搜索旋转排序数组[https://leetcode-cn.c...
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简介:For those who have already solved Search in Rotated Sorted Array, this problem can be solved similarly using codes for that problem and simply adding codes to skip the duplicates. For those who have already solvedSearch in Rotated Sorted Array, this problem can be solved similarly using ...