zoukankan      html  css  js  c++  java
  • 189. 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: [7,1,2,3,4,5,6]
    rotate 2 steps to the right: [6,7,1,2,3,4,5]
    rotate 3 steps to the right: [5,6,7,1,2,3,4]
    

    Example 2:

    Input: [-1,-100,3,99] and k = 2
    Output: [3,99,-1,-100]
    Explanation: 
    rotate 1 steps to the right: [99,-1,-100,3]
    rotate 2 steps to the right: [3,99,-1,-100]

    Note:

    • Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
    • Could you do it in-place with O(1) extra space?

    Approach #1: Extra array. [Java]

    class Solution {
        public void rotate1(int[] nums, int k) {
            int n = nums.length;
            int[] ans = new int[n];
            int start = 0, next = 0;
            while (n--) {
                start = next;
                next = (start + k) % n;
                nums[next] = nums[start];
            }
            for (int i = 0; i < n; ++i) {
                int rev = (i + k) % n;
                ans[rev] = nums[i];
            }
            for (int i = 0; i < n; ++i)
                nums[i] = ans[i];
            return ;
        }
    }
    

      

    Approach #2: Reverse. [Java]

    public class Solution {
        public void rotate(int[] nums, int k) {
            k %= nums.length;
            reverse(nums, 0, nums.length - 1);
            reverse(nums, 0, k - 1);
            reverse(nums, k, nums.length - 1);
        }
        public void reverse(int[] nums, int start, int end) {
            while (start < end) {
                int temp = nums[start];
                nums[start] = nums[end];
                nums[end] = temp;
                start++;
                end--;
            }
        }
    }
    

      

    Approach #3: Cyclic Replacement. [Java]

    public class Solution {
        public void rotate(int[] nums, int k) {
            k = k % nums.length;
            int count = 0;
            for (int start = 0; count < nums.length; start++) {
                int current = start;
                int prev = nums[start];
                do {
                    int next = (current + k) % nums.length;
                    int temp = nums[next];
                    nums[next] = prev;
                    prev = temp;
                    current = next;
                    count++;
                } while (start != current);
            }
        }
    }
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    递增三元子序列
    Linux sed 命令
    linux shell中$0,$?,$!等的特殊用法
    ansible-playbook使用
    Linux下通过crontab命令来实现定时任务
    iperf网络性能测试
    OpenStack接口测试工具rally/tempest环境搭建及使用
    Jenkins配置slaver节点
    微信小程序跳转外部链接(h5页面)以及数据交互
    Echarts 系列之折线图、柱状图相关配置
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10626277.html
Copyright © 2011-2022 走看看