Rotate Array

问题描述

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 ways to solve this problem.

Hint:
Could you do it in-place with O(1) extra space?

算法

代码一

public void rotate(int[] nums, int k) {
int n = nums.length;
int temp;
k = k % n;
for (int i = 0; i < n / 2; i++) {
temp = nums[i];
nums[i] = nums[n - 1 - i];
nums[n - 1 - i] = temp;
}
for (int i = 0; i < k / 2; i++) {
temp = nums[i];
nums[i] = nums[k - 1 - i];
nums[k - 1 - i] = temp;
}
for (int i = 0; i < (n - k) / 2; i++) {
temp = nums[k + i];
nums[k + i] = nums[n - 1 - i];
nums[n - 1 - i] = temp;
}
}

注意k可能会比n大，所以将k模n。

代码二

public void rotate1(int[] nums,int k){
int n=nums.length;
k=k%n;
reverse(nums,0,n-1);
reverse(nums,0,k-1);
reverse(nums,k,n-1);
}
private void reverse(int[] nums,int s,int e){
int temp;
while(s<e){
temp=nums[s];
nums[s]=nums[e];
nums[e]=temp;
s++;
e--;
}
}
private void reverse1(int[] nums,int m,int n){
while(m<n){
nums[m]^=nums
;
nums
^=nums[m];
nums[m]^=nums
;
m++;
n--;
}
}

代码二的思想和代码一的思想一样，优点是把三次反转写成一个函数。另外两个值交换可以用异或运算。

代码三

//https://leetcode.com/discuss/28501/my-c-solution-o-n-time-%26%26-o-1-space
public void rotate2(int[] nums, int k) {
int n = nums.length;
k = k % n;
int i = 0;
int nowIndex = 0, tmp1, tmp2 = nums[0];
for (int j = 0; j < n; j++) {
tmp1 = tmp2;
nowIndex = (nowIndex + k) % n;
tmp2 = nums[nowIndex];
nums[nowIndex] = tmp1;
if (nowIndex == i && i < n - 1) {
i++;
nowIndex = i;
tmp2 = nums[nowIndex];
}

}
}