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...
最直接的办法就是使用两层循环,外层控制旋转次数,内层控制元素交换位置。 publicvoidrotate(int[] nums,intk){if(k % nums.length !=0) {intpre, tem;for(inti=0; i < k % nums.length; i++) { pre = nums[nums.length-1];for(intj=0; j<nums.length; j++) { tem = nums[j]; nums[j]...
leftRotate(arr, 2, 7); printArray(arr, 7); return 0; } Java 实现: class RotateArray { void leftRotate(int arr[], int k, int n) { for (int i = 0; i < k; i++) { leftRotateByOne(arr, n); } } void leftRotateByOne(int arr[], int n) { int temp = arr[0]; for (i...
Find Minimum in Rotated Sorted Array I 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. 二分迭代法 复杂度 时间O(N) 空...
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). You are given a target value to search. If found in the array return its index, otherwise return -1. ...
189. Rotate Array leetcode-java文章分类后端开发 Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. Note: Try to come up as many solutions as you can, there are at...
leftRotate(arr, 2, 7); printArray(arr, 7); return 0; } Java 实现: class RotateArray { void leftRotate(int arr[], int k, int n) { for (int i = 0; i < k; i++) { leftRotateByOne(arr, n); } } void leftRotateByOne(int arr[], int n) { ...
int minNumberInRotateArray(int* nums, int numsLen ) { int min=nums[0]; for (int i=0;ilt;numsLen-1;i++) { if ((nums[i]lt;=nums[i+1])a_牛客网_牛客在手,offer不愁
### 实现代码 ```java #include <stdio.h> void printArray(int arr[], int size); void reverseArray(int arr[], int start, int end); // 将数组左旋 k 个位置 void leftRotate(int arr[], int k, int n) { if (k == 0 || k == n) return; // 防止旋转参数 k 大于数组长度 k =...
Could you do it in-place with O(1) extra space? 要完成的函数: void rotate(vector<int>& nums, int k) 说明: 1、这道题给定一个vector,要求将这个vector最右边的元素调到最左边,重复这个动作k次,最终结果仍然存放在nums中。要求空间复杂度为O(1)。