leetcode-javscript-189. 旋转数组

输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]

旋转法 逆转法看不懂
暴力法

public class Solution {
    public void rotate(int[] nums, int k) {
        int temp, previous;
        for (int i = 0; i < k; i++) { //三次k
            previous = nums[nums.length - 1];//取三次最后面的元素
            for (int j = 0; j < nums.length; j++) { //nusmlength次
                temp = nums[j];//位置交换
                nums[j] = previous;
                previous = temp; //nums[nums.length = nums[j[]
            }
        }
    }
}

额外数组

public class Solution {
    public void rotate(int[] nums, int k) {
        int[] a = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            a[(i + k) % nums.length] = nums[i]; // 超过length 开始取0 1 2
        }
        for (int i = 0; i < nums.length; i++) {
            nums[i] = a[i];
        }
    }
}

环装替换
方法 3：使用环状替换
算法

image.png

nums: [1, 2, 3, 4, 5, 6]
k: 2

换状替换

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);
        }
    }
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/rotate-array/solution/xuan-zhuan-shu-zu-by-leetcode/
来源：力扣（LeetCode）
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