Array-Rotate Array

最新推荐文章于 2021-08-03 10:47:58 发布

原创最新推荐文章于 2021-08-03 10:47:58 发布 · 290 阅读

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本文介绍了将数组元素向右旋转k步的三种解决方案，包括一种简单但空间复杂度较高的方法，一种空间复杂度为O(1)的更优方案，以及最简洁高效的实现方式。

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Description:

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.

My Solution:

class Solution {
//缺点在于空间复杂度为O(n)，使用了temp这个额外存储空间。
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        int[] temp = new int[len];
        for(int i = 0;i < len;i++){
            temp[(i + k) % len] = nums[i];
        }
        for(int i = 0;i < len;i++){
            nums[i]  = temp[i];
        }
    }
}

Better Solution:

class Solution {     
    //优点在于空间复杂度为O(1)
    public void rotate(int[] nums, int k) {
        if(nums == null || nums.length < 2 || k == 0) return;
        if(k > nums.length) k = k%nums.length;
        int rot = nums.length -1 -k;
        reverse(0, rot, nums);
        reverse(rot+1, nums.length-1, nums);
        reverse(0, nums.length-1, nums);
    }

    public int[] reverse(int start, int end, int[] nums) {
        if(nums == null || nums.length<2) return nums;
        while(start < end) {
            int tmp = nums[start];
            nums[start] = nums[end];
            nums[end] = tmp;
            start++;
            end--;
        }


        return nums;
    }
}

Best Solution:

//把nums从前后分成两截，调换一下位置，非常简洁
class Solution {
    public void rotate(int[] nums, int k) {
        k=k%nums.length;
        int[] part1= Arrays.copyOf(nums,nums.length-k);
        int[] part2=Arrays.copyOfRange(nums,nums.length-k,nums.length);
        System.arraycopy(part2,0,nums,0,part2.length);
        System.arraycopy(part1,0,nums,part2.length,part1.length);
    }
}