boolresult=searchD(nums, target, 0, size-1); returnresult; } };
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...
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...
There is an integer array nums sorted in ascending order (with distinct values).Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1]
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