题解 | #旋转数组#
旋转数组
https://www.nowcoder.com/practice/e19927a8fd5d477794dac67096862042
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# 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
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# 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
