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