题目链接:https://leetcode.com/problems/find-minimum-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). Find the minimum element. You may assume no duplicate exists in the array. 思...
Suppose a sorted array 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). Find the minimum element. You may assume no duplicate exists in the array. SOLUTION 1: 这个题目trick的地方在于,它的旋转pivot次数未知。所以有可能它转...
一、题目描述 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). Find the minimum element. You may assume no duplicate exists in the array. 二、分析 这题难度有,因为他默认的是数组中所有的元素是不同的。只...
153. Find Minimum in Rotated Sorted Array 题目 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]). Find the min element. You may assume no duplicate exists in the array. Example ...
给定一个特殊升序数组,即一个排序好的数组,把前边的若干的个数,一起移动到末尾,找出最小的数字。 解法一 其实之前已经在 33 题 解法一中写过这个解法了,这里直接贴过来。 求最小值,遍历一遍当然可以,不过这里提到了有序数组,所以我们可以采取二分的方法去找,二分的方法就要保证每次比较后,去掉一半的元素。 这...
Can you solve this real interview question? Find Minimum in Rotated Sorted Array - Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: * [4,5,6,7,0,1,2] if
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Given the sorted rotated array nums of unique elements, return the minimum element of this array. ...
153 Find Minimum in Rotated Sorted Array "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]). Find the minimum element. ...
importjava.util.Scanner;publicclassFindMinimumElementInRotatedSortedArray{privatestaticintfindMinimumElement(int[]a){intn=a.length;intstart=0;intend=n-1;// If the first element is less than the last element then there is no rotation. The first element is minimum.if(a[start]<=a[end]){return...
#include<stdio.h>// Function to find the minimum element in a rotated sorted arrayintfindMin(intarr1[],intstart,intend){// If the start index and end index are the same, return the element at that indexif(start==end){returnarr1[start];}// Find the middle indexintmid=(start+end)...