leetcode 189. Rotate Array-旋转数组

原题链接：189. Rotate Array

【思路1-Java、Python】时间复杂度O(n)，空间复杂度O(1)

以[1,2,3,4,5,6,7], k=3 为例，经过3次翻转，就能得到结果：

        reverse(nums, 0, nums.length - 1);  //[7,6,5,4,3,2,1]        reverse(nums, 0, k - 1);            //[5,6,7,4,3,2,1]        reverse(nums, k, nums.length - 1);  //[5,6,7,1,2,3,4]

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);    }    private void reverse(int[] nums, int m, int n) {        while (m < n) {            int temp = nums[m];            nums[m++] = nums[n];            nums[n--] = temp;        }    }}

33 / 33 test cases passed. Runtime: 1 ms  Your runtime beats 11.70% of javasubmissions.

class Solution(object):    def rotate(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: void Do not return anything, modify nums in-place instead.        """        k %= len(nums)        self.reverse(nums, 0, len(nums) - 1)        self.reverse(nums, 0, k - 1)        self.reverse(nums, k, len(nums) - 1)    def reverse(self, nums, start, end) :        while start < end :            nums[start], nums[end] = nums[end], nums[start]            start += 1            end -= 1

33 / 33 test cases passed. Runtime: 84 ms  Your runtime beats 49.51% of pythonsubmissions.

【思路2-Java、Python】时间复杂度O(n)，空间复杂度O(1)

以 k 为长度，循环移动 n 次，

以[1, 2, 3, 4, 5, 6, 7], k = 3为例，过程是这样的：

[1, 2, 3, 1, 5, 6, 7]

[1, 2, 3, 1, 5, 6, 4]

[1, 2, 7, 1, 5, 6, 4]

[1, 2, 7, 1, 5, 3, 4]

[1, 6, 7, 1, 5, 3, 4]

[1, 6, 7, 1, 2, 3, 4]

[5, 6, 7, 1, 2, 3, 4]

[5, 6, 7, 1, 2, 3, 4]

然而有时候有可能循环移动会出现死循环情况，所以要加上①处判断条件，以[1, 2, 3, 4, 5, 6], k = 2为例

[1, 2, 1, 4, 5, 6]

[1, 2, 1, 4, 3, 6]

[5, 2, 1, 4, 3, 6]

[5, 2, 1, 4, 3, 6]  //这时 distance = 0了，所以要调整

[5, 2, 1, 2, 3, 6]

[5, 2, 1, 2, 3, 4]

[5, 6, 1, 2, 3, 4] //同样 distance = 0

[5, 6, 1, 2, 3, 4]

public class Solution {    public void rotate(int[] nums, int k) {        for (int i=0, idx=0, cur=nums[0], distance=0, n=nums.length; i < n; i++) {            idx = (idx + k) % n;                        int temp = cur;  //交换 cur 和 nums[idx]            cur = nums[idx];            nums[idx] = temp;                        distance = (distance + k) % n;  //①            if (distance == 0) {                idx = (idx + 1) % n;                cur = nums[idx];            }        }    }}

33 / 33 test cases passed. Runtime: 2 ms  Your runtime beats 2.72% of javasubmissions.

class Solution:    # @param nums, a list of integer    # @param k, num of steps    # @return nothing, please modify the nums list in-place.    def rotate(self, nums, k):        n = len(nums)        idx = 0        distance = 0        cur = nums[0]        for x in range(n):            idx = (idx + k) % n            nums[idx], cur = cur, nums[idx]                        distance = (distance + k) % n            if distance == 0:                idx = (idx + 1) % n                cur = nums[idx]

33 / 33 test cases passed. Runtime: 88 ms  Your runtime beats 40.12% of pythonsubmissions.