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] <= target < nums[mid],说...
Can you solve this real interview question? Search in Rotated Sorted Array II - There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values). Before being passed to your function, nums is rotated at an unknown piv
初始版本 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). You are given a target value to search. If found in the array return its index, otherwise return -1. You ma...
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. ...
Write a function to determine if a given target is in the array. The array may contain duplicates. 这道题很简单,直接遍历即可。也可以使用二分查找。 代码如下: AI检测代码解析 public class Solution { /* * 最简单的就是遍历 * */ public boolean search(int[] nums, int target) ...
There is an integer arraynumssorted in ascending order (withdistinctvalues). Prior to being passed to your function,numsispossibly rotatedat an unknown pivot indexk(1 <= k < nums.length) such that the resulting array is[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ...
publicintsearch(int[]nums,inttarget){intlo=0,hi=nums.length-1;while(lo<=hi){intmid=lo+(hi-lo)/2;intnum=nums[mid];//nums [ mid ] 和 target 在同一段if((nums[mid]<nums[0])==(target<nums[0])){num=nums[mid];//nums [ mid ] 和 target 不在同一段,同时还要考虑下变成 -inf 还...
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在经过了财务比赛的洗礼后,我还是回归到了日复一日的刷题模式当中。考虑到可能会看我文章的读者在阅读中文时会更加轻松,从这篇文章开始LeetCode的题目将以中文形式显示。 今天的笔记包含基于改造二分法(Modified Binary Search)类型下的7个题目,它们在leetcode上的编号和题名分别是: ...