leetcode--Rotate Array

最新推荐文章于 2025-12-19 16:34:09 发布

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本文介绍了数组旋转问题的三种解决方案，包括直接法O(n)、反转法O(1)及两次循环O(1)。每种方法都有详细的代码实现，适合初学者理解和实践。

题目：Rotate Array

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.

数组的旋转平移

One:O(n) 直接

public class Solution {
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        int[] res = new int[len];
        k %=len;
        for(int i=0;i<len;i++){
            res[(i+k)%len]=nums[i];
        }
        for(int i=0;i<len;i++){
            nums[i] = res[i];
        }
    }
}

Two: O(1)+reverse()

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

Three:两次循环O(1)

public class Solution {
    public void rotate(int[] nums, int k) {
       if (nums.length == 0) return;
    int n = nums.length;
    while ((k %= n) > 0 && n > 1) {
        int range = n - k;
        for (int i = 1; i <= range; i++) {
            int val = nums[n - i];
            nums[n - i] = nums[n - i - k];
            nums[n - i - k] = val;
        }
        n = k;
        k = n - (range % k);
    }
    } 
}