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[n];
            nums[n]^=nums[m];
            nums[m]^=nums[n];
            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];
            }

        }
    }