Question

我在leetcode中读到了这样的问题

给出一个数组，将数组向右旋转 k 步，其中 k 为非负数。

示例1：
Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
示例2：
Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
注意：


尝试提供尽可能多的解决方案，至少有3种不同的方法来解决此问题。

您可以在原位使用O（1）多余的空间吗？

它的官方解决方案是用Java编写的

public class Solution {
    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        int count = 0;
        for (int start = 0; count < nums.length; start++) {
            int current = start;
            int prev = nums[start];
            do {
                int next = (current + k) % nums.length;
                int temp = nums[next];
                nums[next] = prev;
                prev = temp;
                current = next;
                count++;
            } while (start != current);
        }
    }
}

我试图将其翻译为python

class Solution3:
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        k = k % len(nums)
        count = 0
        start = 0
        while count < len(nums):
            current = start 
            prev = nums[start] #store the value in the position

            while start != current:
                next = (current + k) % len(nums)
                temp = nums[next]
                nums[next] = prev 
                prev = temp #store the value 
                current = next 
                count += 1

但是，没有do while

这样的逻辑

如何处理这种情况？

Answer 1

在Python中，您可以像这样模拟do while的行为

while True:
    do_things()
    if not cond:
        break

所以在您的情况下：

while True:
    next = (current + k) % len(nums)
    temp = nums[next]
    nums[next] = prev 
    prev = temp #store the value 
    current = next 
    count += 1

    if start == current:
        break

模仿在执行逻辑控制时处理旋转问题的方法

1 个答案: