题解 | #旋转数组#

旋转数组

https://www.nowcoder.com/practice/e19927a8fd5d477794dac67096862042

#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# 旋转数组
# @param n int整型 数组长度
# @param m int整型 右移距离
# @param a int整型一维数组 给定数组
# @return int整型一维数组
#
class Solution:
    def solve(self , n: int, m: int, a: List[int]) -> List[int]:
        # write code here
        moved = n
        m = m-(m//n)*n
        head = 0
        idx = 0
        cur = a[head-m]

        while m and moved:
            if m >1 and n%m ==0 and n/2 == moved:
                head += 1
                cur = a[head - m]
            pre = a[(head+idx)%n]
            a[(head+idx)%n] = cur
            cur = pre
            idx += m
            idx %= n
            moved -= 1

        return a

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