classRotateArray{// 将数组 arr 向左旋转 k 个位置voidleftRotate(intarr[],intk,intn){// 处理 k >= n 的情况,比如 k = 13, n = 12k = k % n;inti, j, s, temp;// s = j + k;intgcd=gcd(k, n);for(i =0; i < gcd; i++) {// 第 i 轮移动元素temp = arr[i]; j =...
1publicvoidrotateByOne(int[] arr) {2if(arr ==null|| arr.length == 0){3return;4}5inttemp = arr[arr.length - 1];6for(inti = arr.length - 1; i > 0; i--){7arr[i] = arr[i - 1];8}9arr[0] =temp;10} Follow up question: instead of rotating the given array by one po...
Input: N = 5, array[] = {1,2,3,4,5} Output: 2,3,4,5,1 Explanation: Since all the elements in array will be shifted toward left by one so ‘2’ will now become the first index and and ‘1’ which was present at first index will be shifted at last. Example 2: Input: N ...
将array2的后n - k(n为数组元素个数)个元素前后颠倒得到array3,即最终结果 Java代码实现如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{publicvoidrotate(int[]nums,int k){k%=nums.length;reverse(nums,0,nums.length-1);reverse(nums,0,k-1);reverse(nums,k,nums.length-1);...
LeetCode笔记:189. Rotate Array 大意: 旋转一个有n个元素的数组,右移k步。 比如,n = 7,k = 3,数组 [1,2,3,4,5,6,7] 就被旋转为 [5,6,7,1,2,3,4]。 思路: 旋转本身没太多特别好说的,我的做法是用另一个数组来记录旋转后的内容,然后复制回原数组。当然记录时是从第nums.length-k个元素...
Rotate Array 这题当然有很朴素的解法,例如利用k % nums.length次循环每次循环都将原字符串向右推移1位,或者直接计算出每个字符最终所应该在的位置直接进行赋值。 那么这两种方法,前者复杂度太高,或者不够清晰简单。 我选用的是在编程珠玑中提到的翻转方法,比如我们的输入是[1,2,3,4,5,6,7]和k = 3,那么...
Leetcode: Rotate Array 题目: 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 least 3 different...
Rotate Array Given an array, rotate the array to the right by k steps, where k is non-negative. Example 1: Input: [1,2,3,4,5,6,7] and k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,......
189 Rotate Array 旋转数组 将包含 n 个元素的数组向右旋转 k 步。 例如,如果 n = 7 , k = 3,给定数组 [1,2,3,4,5,6,7] ,向右旋转后的结果为 [5,6,7,1,2,3,4]。 注意: 尽可能找到更多的解决方案,这里最少有三种不同的方法解决这个问题。
Given an array, rotate the array to the right byksteps, wherekis non-negative. We need to master the 4 ways to solve this problem, especially the last two methods: 1. Brute Force:move the array to the rightktimes 2. Another Array:move each element in the array to the final position ...