代码//LeetCode, Search in Rotated Sorted Array II//时间复杂度 O(n),空间复杂度 O(1)classSolution {public:boolsearch(intA[],intn,inttarget) {intfirst =0, last =n;while(first !=last) {constintmid = first + (last - first)
1. nums[mid] = target,则可以返回mid 2. nums[mid] < nums[right],说明在[mid, right]区间是右边递增的区间,然后判断target是否在这个区间内 1)如果nums[mid] < target <= nums[right],说明target在右边区间里,则left = mid + 1; 2)否则在左边区间里,搜索左边区间,right = mid - 1; 3. nums[m...
Example 1: Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true Example 2: Input: nums = [2,5,6,0,0,1,2], target = 3 Output: false Noted that: The array may contain dumplicated value, so this is different to Leet Code 33. Search in Rotated Sorted Array In the ...
This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 描述 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。 请找出其中最小的...
You may assume no duplicate exists in the array. 就是在一个旋转后的数组中寻找target数据。 很明显,最简单的方法就是遍历一次,不过题目的意思明显不是要求这么做的,主要考察的还是使用二分查找,所以这里使用二分查找来做。 代码如下: public class Solution { ...
LeetCode: 154. Find Minimum in Rotated Sorted Array II 题目描述 Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). ...
小范的leetcode日记(33)Search in Rotated Sorted Array, 视频播放量 16、弹幕量 0、点赞数 1、投硬币枚数 0、收藏人数 1、转发人数 0, 视频作者 小范刷题, 作者简介 ,相关视频:小范的leetcode日记(81)Search in Rotated Sorted Array II,小范的leetcode日记(80)Remove D
Can you solve this real interview question? Find Minimum in Rotated Sorted Array - Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: * [4,5,6,7,0,1,2] if
1. Description 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++;cont...
寻找旋转排序数组中的最小值 II 思路解题方法复杂度 Code 思路该方法采用二分查找的变种,逐步缩小搜索范围,最终找到数组中的最小值解题方法 如果 nums[mid] 等于 nums[right],这意味着 mid 到 right 之间的元素可能存在重复值。通过将 right 二分查找 数组 Python3 1 100 0 C++ 智能模式 1 2 3 4 5 6...