189. 轮转数组

https://leetcode.cn/problems/rotate-array/description/

给定一个整数数组 nums,将数组中的元素向右轮转 k 个位置,其中 k 是非负数。

示例 1:

1
2
3
4
5
6
输入: nums = [1,2,3,4,5,6,7], k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右轮转 1 步: [7,1,2,3,4,5,6]
向右轮转 2 步: [6,7,1,2,3,4,5]
向右轮转 3 步: [5,6,7,1,2,3,4]

示例 2:

1
2
3
4
5
输入:nums = [-1,-100,3,99], k = 2
输出:[3,99,-1,-100]
解释:
向右轮转 1 步: [99,-1,-100,3]
向右轮转 2 步: [3,99,-1,-100]

提示:

  • 1 <= nums.length <= 105
  • -231 <= nums[i] <= 231 - 1
  • 0 <= k <= 105

进阶:

  • 尽可能想出更多的解决方案,至少有 三种 不同的方法可以解决这个问题。
  • 你可以使用空间复杂度为 O(1)原地 算法解决这个问题吗?

解法一:额外数组

1
2
3
4
5
6
7
8
9
class Solution {
public void rotate(int[] nums, int k) {
int[] newNums = new int[nums.length];
for (int i = 0; i < newNums.length; i++) {
newNums[(i + k) % nums.length] = nums[i];
}
System.arraycopy(newNums, 0, nums, 0, nums.length);
}
}

解法二:反转数组

image-20230802133347557

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
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 += 1;
end -= 1;
}
}
}

解法三:环状替换(实在想不到了)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
public void rotate(int[] nums, int k) {
int n = nums.length;
k = k % n;
int count = gcd(k, n);
for (int start = 0; start < count; ++start) {
int current = start;
int prev = nums[start];
do {
int next = (current + k) % n;
int temp = nums[next];
nums[next] = prev;
prev = temp;
current = next;
} while (start != current);
}
}

public int gcd(int x, int y) {
return y > 0 ? gcd(y, x % y) : x;
}
}

作者:力扣官方题解
链接:https://leetcode.cn/problems/rotate-array/solutions/551039/xuan-zhuan-shu-zu-by-leetcode-solution-nipk/
来源:力扣(LeetCode)
0%