publicclassSolution {/*** param A : an integer ratated sorted array and duplicates are allowed * param target : an integer to be search * return : a boolean*/publicbooleansearch(int[] A,inttarget) {if(A ==null|| A.length == 0)returnfalse;intlb = 0, ub = A.length - 1;while(...
版本2仍然work的原因是,当mid靠到Left这边时,left的值与mid相同,我们这时left++就丢弃了不可用的值,所以这个算法没有问题。 LeetCode: Search in Rotated Sorted Array 解题报告- Yu's ...中就不可以这样了,判断是否有序时,必须使用<=,因为题1中没有第三 个分支:直接跳过。 View Code SOLUTION 2: 1. ...
81. Search in Rotated Sorted Array II There is an integer arraynumssorted in non-decreasing order (not necessarily withdistinctvalues). Before being passed to your function,numsisrotatedat an unknown pivot indexk(0 <= k < nums.length) such that the resulting array is[nums[k], nums[k+1...
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). 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。...
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? 分析 题目的意思是:给你一个旋转的有序数组,数组中是否存在一个数等于给定的target。
[ 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?
leetCode 33. Search in Rotated Sorted Array(c语言版本) bingo酱 I am a fighter 来自专栏 · leetcode每日斩 题目大意: 给定一个旋转升序的数组,所谓旋转就是假设把头和尾连接起来,然后找到最小那个数开始,往后开始就是升序的,直到再回到最小那个数为止。这样看起来就像一个环一样。 然后,现在给定一个数,...
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...
This is a follow up problem toSearch in Rotated Sorted Array, wherenumsmay contain duplicates. Would this affect the run-time complexity? How and why? 代码: BinarySearch方法: classSolution:defsearch(self,nums,target):""" :type nums: List[int] ...
The simplest solution would be to check every element one by one and compare it with k (a so-called linear search). This approach works in O(n) , but doesn't utilize the fact that the array is sorted.Binary search of the value $7$ in an array. The image ...