zoukankan      html  css  js  c++  java
  • [GeeksForGeeks] Cyclically rotate an array by one

    Given an array of integers, write a program that cyclically rotates the array by one.

    Example: given {1,2,3,4,5}, your program should rotate the array to {5,1,2,3,4}

    Algorithm:

    1. save the last element in a temp variable.

    2. starting from the last element until the second element, copy its previous element to be their values.

    3. copy temp to the first element.

    O(n) runtime, O(1) space

     1 public void rotateByOne(int[] arr) {
     2     if(arr == null || arr.length == 0){
     3         return;
     4     }
     5     int temp = arr[arr.length - 1];
     6     for(int i = arr.length - 1; i > 0; i--){
     7         arr[i] = arr[i - 1];
     8     }
     9     arr[0] = temp;
    10 }

    Follow up question:  instead of rotating the given array by one position, rotate it by m positions, m should be controlled by 

    your program caller. if m is positive, it means rotating right, if m is negative, it means rotating left.

    Solution 1. O(m * n) runtime

    Simply call rotateByOne m times.

    Solution 2. O(n) runtime, O(1) space

    Algorithm:  Acheive the rotation by swaping values only one time instead of  m times for each location.

    1. Move arr[0] to a temporary variable t, then move arr[m] to arr[0], arr[2 * m] to arr[m], and so on.(taking all indices into arr modulo arr.length), until we come back to taking the element from arr[0], 

        at which point we instead take t and stop the process.

    2. If step 1 does not move all the elements, repeat step 1 starting from arr[1].  Repeat this step until all the elements are moved, advancing the start element to the next location each iteration.

    The following code only shows the left rotation implementation, the right counter part is similar.

     1 public static void leftRotateByM(int[] arr, int m) {
     2     if(m % arr.length != 0) {
     3         int count = 0;
     4         int startIdx = 0;                   
     5         while(count < arr.length) {
     6             int currIdx = startIdx;
     7             int nextIdx = (currIdx + m) % arr.length;
     8             int temp = arr[startIdx];
     9             while(nextIdx != startIdx) {
    10                 arr[currIdx] = arr[nextIdx];
    11                 currIdx = nextIdx;
    12                 nextIdx = (nextIdx + m) % arr.length;
    13                 count++;
    14             }
    15             arr[currIdx] = temp;
    16             startIdx++;
    17             count++;
    18         }
    19     }
    20 }

    Solution 3. O(n) runtime, O(1) space, using reverse algorithm

    1. decide on the split point depending if it is a left or right rotation.

    2. reverse the splitted array separately, then reverse the whole array.

     1 private void reverse(int[] arr, int start, int end) {
     2     while(start < end) {
     3         arr[start] ^= arr[end];
     4         arr[end] ^= arr[start];
     5         arr[start] ^= arr[end];
     6         start++;
     7         end--;
     8     }
     9 }
    10 public void rotateByMUsingReverse(int[] arr, int m) {
    11     if(arr == null || arr.length == 0 || m % arr.length == 0){
    12         return;
    13     }
    14     m %= arr.length;
    15     int splitIdx = m > 0 ? arr.length - m : m;
    16     reverse(arr, 0, splitIdx - 1);
    17     reverse(arr, splitIdx, arr.length - 1);
    18     reverse(arr, 0, arr.length - 1);
    19 }
  • 相关阅读:
    docker学习记录
    TCP/IP基础介绍
    JS对select操作
    js中删除table里所有行
    端口
    js中定时器的使用
    ASP.NET程序中常用的三十三种代码
    NHibernate学习(转)
    条面向对象设计的经验原则(转)
    客户端等select和input控件
  • 原文地址:https://www.cnblogs.com/lz87/p/7348678.html
Copyright © 2011-2022 走看看