Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 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 duplicate exists in the array...
LeetCode系列(六)- Search in Rotated Sorted Array 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])。 搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回-1。 你可以假设数组中不存在重复的元素。 你的算法...
题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array/题目: Suppose a sorted array is rotated at some pivot unknown to you beforehand. 0 1 2 4 5 6 7might become4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, o...
LeetCode-33-搜索旋转排序数组(Search in Rotated Sorted Array) 33. 搜索旋转排序数组[https://leetcode-cn.c...
输出: 4 示例2 输入: nums = [4,5,6,7,0,1,2], target = 3 输出: -1 解题思路: 主要是利用二分查找法来解题 因为原始数组是升序数组,所以旋转之后的数组必是两个升序数组的合并 首先先通过一次二分查找的方式,找到第二个升序数组的第一项l,即有序数组的最小一项 ...
Leetcode 33. Search in Rotated Sorted Array Search in Rotated Sorted Array 题目大意:给定一个有序数组,经过旋转得到新得数组,也即改变起点位置,如[0,1,2,4,5,6,7] 变成[4,5,6,7,0,1,2]。要求仍然在log的时间进行查找某个特定的target...
此时medium = 3,正中间的数是0,根据情况2的提示,我们先判断到nums[low] = 4 是比nums[medium] = 0大的,说明旋转升序的起始点,这里是位置3,包含这low到medium之间,所以才会出现这种情况,同时这种情况也说明了,从medium到high之间的数,肯定是升序的,即Subnums2 = nums[medium to high]是升序数组,如果nums[...
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 ...
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. 二分法 思路 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。假设数组是A,每次左边缘为l,右边缘为r,还有中间位...
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], ...