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[mid] > nums[right],说明[elft, mid]区间是在左边的递增区间,然后判断target是否在这个左边区间里 1)如果nums[left] <= targe
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. 解法:同旋转数组查找I,本题也使用二分查找,只是在nums[mid]和nums[right](或者nums[right])...
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 in the array return true, otherwise return false. Example 1: Input: num...
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. ...
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
[ 4 5 6 7 1 2 3] ,如果 target = 2,那么数组可以看做 [ -inf -inf - inf -inf 1 2 3]。 和解法一一样,我们每次只关心 mid 的值,所以 mid 要么就是 nums [ mid ],要么就是 inf 或者 -inf。 什么时候是 nums [ mid ] 呢?
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...
https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ 解题思路 先找到数组被rotated的位置如果有的话。 确定好位置之后再在排序的数据区间内查找元素。 代码 class Solution{public:boolsearch(vector<int>&nums,inttarget){if(nums.empty()){returnfalse;}//先查找被反转的位置intpos=findPos(...
# Special considerationsiflen(matrix)==0:returnFalseiflen(matrix[0])==0:returnFalserow=len(matrix)column=len(matrix[0])left=0right=row*column-1whileleft<=right:mid=left+(right-left)//2# Get the position using math (attention: when converting 1-D array back to 2-D coordinate,# the ...
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. 因为rotate的缘故,当我们切取一半的时候可能会出现误区,所以我们要做进一步的判断。具体来说,假...