給定一個(gè)數(shù)組，將數(shù)組中的元素向右移動(dòng) k 個(gè)位置，其中 k 是非負(fù)數(shù)。
示例 1:
輸入: [1,2,3,4,5,6,7] 和 k = 3
輸出: [5,6,7,1,2,3,4]
解釋:
向右旋轉(zhuǎn) 1 步: [7,1,2,3,4,5,6]
向右旋轉(zhuǎn) 2 步: [6,7,1,2,3,4,5]
向右旋轉(zhuǎn) 3 步: [5,6,7,1,2,3,4]
示例 2:
輸入: [-1,-100,3,99] 和 k = 2
輸出: [3,99,-1,-100]
解釋:
向右旋轉(zhuǎn) 1 步: [99,-1,-100,3]
向右旋轉(zhuǎn) 2 步: [3,99,-1,-100]

解法一：開(kāi)辟新的數(shù)組

n = len(nums)
Array = [0]*n

for i in range(n):
    Array[(i+k)%n] = nums[i]
nums[:]=Array

解法二插入法

k = k%len(nums)
for i in range(k):
    nums.insert(0,nums.pop())

解法三數(shù)組翻轉(zhuǎn)

nums[:] = nums[::-1]
nums[:k] = nums[:k][::-1]
nums[k:][:] = nums[k:][::-1]

補(bǔ)充：雙指針?lè)崔D(zhuǎn)數(shù)組

def Reverse(nums):
    left = 0
    right = len(nums)-1
    while left<right:
        nums[left],nums[right] = nums[right],nums[left]
        left+=1
        right+=1

解法四

def gcd(m, n):
    if m > n:
        m, n = n, m
    while m > 0:
        mod = n % m
        n, m = m, mod
    return n

n = len(nums)
count = gcd(n, k)

for i in range(count):
    cur = i
    prev = nums[i]
    while True:
        last = (cur + k) % n
        temp = nums[last]
        nums[last] = prev
        #prev記錄上一個(gè)元素
        prev = temp
        cur = last
        if cur == i:
            break