189. Rotate Array

题目：

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.

[show hint]

Related problem: Reverse Words in a String II

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

 

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链接：  http://leetcode.com/problems/rotate-array/

题解：

数组移动问题。可以参考<编程珠玑>里利用三次reverse达成数组向左移动的方法。向右移动k step相当于向左移动数组长度nums.length - k。 要注意提前取摸，使k成为一个合理的值。

Time Complexity - O(n)， Space Complexity - O(1)。

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

 

Update:

依然是三步反转法，先反转整个数组，再翻转0 ~ k - 1， 最后翻转 k ~ n - 1

public class Solution {
    public void rotate(int[] nums, int k) {
        if(nums == null || nums.length == 0)
            return;
        int n = nums.length;
        k %= n;
        
        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);
    }
    
    private void reverse(int[] nums, int lo, int hi) {
        while(lo < hi)
            swap(nums, lo++, hi--);
    }
    
    private void swap(int[] nums, int lo, int hi) {
        int tmp = nums[lo];
        nums[lo] = nums[hi];
        nums[hi] = tmp;
    }
}

 

二刷: 

三步反转法.

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k == 0) {
            return;
        }       
        int len = nums.length;
        k %= len;
        reverse(nums, 0, len - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, len - 1);
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    } 
    
    private void reverse(int[] nums, int i, int j) {
        while (i < j) {
            swap(nums, i++, j--);
        }
    }
}

 

三刷:

Java:

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k <= 0) return;
        int len = nums.length;
        k %= len;
        reverse(nums, 0, len - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, len - 1);
    }
    
    private void reverse(int[] nums, int i, int j) {
        while (i < j) {
            int tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
            i++;
            j--;
        }
    }
}

 

 

Reference: