You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no duplicate exists in the array. 做完了Find Minimum in Rotated Sorted Array之后,对这题又发现了一种非常简单的做法。 Find Minimum in Rotated Sorted Array的思路如下: 当nums...
RSA:旋转排序数组,即Rotated Sorted Array 样例1: 输入:数组 = [4, 5, 1, 2, 3],target = 1 输出:2 解释: 元素1在数组中对应索引位置为2。 样例2: 输入:数组 = [4, 5, 1, 2, 3],target = 0 输出:-1 解释: 元素0不在数组中,返回-1。 3 思路 因为RSA是分段有序...
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...
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 may assume no...
81. Search in Rotated Sorted Array II https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ 这两道题都是属于循环有序数组的查找问题,无论是查找具体元素的下标还是检查一个目标值是否在数组中,主要的判断逻辑都是一致的。
33. Search in Rotated Sorted Array 题解 题目意思就是在一个上升序列中截断成AB 两个子序列 然后BA拼接成一个新的序列, 而题目要求 我们查找在这个序列中是否存在target,如果看到这里就会发现 这个很简单啊,查找一个序列中是否有某个数字 直接遍历就好了 方法木佬佬,但是 最重要的是题目末尾有一句 ...
There is an integer array nums sorted in ascending order (with distinct values). 一个升序排序的整数数组 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], ......
1. A computerized method comprising at a server: storing a plurality of expert profiles in a database, each of the plurality of expert profiles comprising information associated with a person having knowledge in a particular category, subject or topic; receiving search criteria over a network from...
// Search in Rotated Sorted Array II// Time Complexity: O(n),Space Complexity: O(1)classSolution{public:boolsearch(constvector<int>&nums,inttarget){intfirst=0,last=nums.size();while(first!=last){constintmid=first+(last-first)/2;if(nums[mid]==target)returntrue;if(nums[first]<nums[...
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. ...