标签：bsp runtime geeks follow private solution update var cal
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. pick one element and save its value in a temp variable.
2. copy the new value from the source location to this destination location. Update destination location
to the source location; Repeat this process until there is only one element left that has not been correctly
updated. Set this element‘s value to temp.
1 public class Solution { 2 private void rightRotateByOne(int[] arr) { 3 int temp = arr[arr.length - 1]; 4 for(int i = arr.length - 1; i > 0; i--){ 5 arr[i] = arr[i - 1]; 6 } 7 arr[0] = temp; 8 } 9 private void leftRotateByOne(int[] arr) { 10 int temp = arr[0]; 11 for(int i = 0; i < arr.length - 1; i++){ 12 arr[i] = arr[i + 1]; 13 } 14 arr[arr.length - 1] = temp; 15 } 16 private void rightRotateByM(int[] arr, int m) { 17 //O(m * n) runtime 18 /*for(int i = 0; i < m; i++){ 19 rightRotateByOne(arr); 20 }*/ 21 //O(n) runtime 22 if(m % arr.length != 0) { 23 m %= arr.length; 24 int count = 0; 25 int currIdx = arr.length - 1; 26 int temp = arr[arr.length - 1]; 27 while(count < arr.length - 1){ 28 arr[currIdx] = arr[(currIdx - m + arr.length) % arr.length]; 29 currIdx = (currIdx - m + arr.length) % arr.length; 30 count++; 31 } 32 arr[currIdx] = temp; 33 } 34 } 35 private void leftRotateByM(int[] arr, int m) { 36 //O(m * n) runtime 37 /*for(int i = 0; i < m; i++){ 38 leftRotateByOne(arr); 39 }*/ 40 //O(n) runtime 41 if(m % arr.length != 0) { 42 m %= arr.length; 43 int count = 0; 44 int currIdx = 0; 45 int temp = arr[0]; 46 while(count < arr.length - 1){ 47 arr[currIdx] = arr[(currIdx + m) % arr.length]; 48 currIdx = (currIdx + m) % arr.length; 49 count++; 50 } 51 arr[currIdx] = temp; 52 } 53 } 54 public void rotateByM(int[] arr, int m) { 55 if(arr == null || arr.length == 0 || m % arr.length == 0){ 56 return; 57 } 58 if(m > 0) { 59 rightRotateByM(arr, m); 60 } 61 if(m < 0) { 62 leftRotateByM(arr, Math.abs(m)); 63 } 64 } 65 }
[GeeksForGeeks] Cyclically rotate an array by one
标签：bsp runtime geeks follow private solution update var cal
原文地址：http://www.cnblogs.com/lz87/p/7348678.html
