这是我再 leetcode 上遇见的一道题: Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
我是这样完成的
# @param {Integer[]} nums
# @param {Integer} k
# @return {void} Do not return anything, modify nums in-place instead.
def rotate(nums, k)
k %=nums.length
if k!= 0
t = nums[k..(nums.length-1).to_i]
f = nums[0..(k-1).to_i]
nums = t + f
end
end
问题是 nums 的值只在函数内改变了
serco
#1
2015年04月01日
k 允许大于 n 吗？可以用 Ruby 内建方法吗？
def rotate(nums, k)
head = nums.shift(k%m)
nums.push(*head)
end
nums.instance_exec(k) do
k %=self.length
if k!= 0
t = self[k..(self.length-1).to_i]
f = self[0..(k-1).to_i]
nums = t + f
end
end
作用域问题
def xxx nums,k
nums.instance_exec(k) do
k %=self.length
if k!= 0
t = self[k..(self.length-1).to_i]
f = self[0..(k-1).to_i]
nums = t + f
end
end
end
nums = [1,2]
xxx nums,1
p nums
.......................Sub_table/sources/a.rb" [1, 2]
Process finished with exit code 0
这样使用就是说你不需要在定义方法了，直接使用：
nums = [1,2]
k = 1
nums.instance_exec(k) do
k %=self.length
if k!= 0
t = self[k..(self.length-1).to_i]
f = self[0..(k-1).to_i]
nums = t + f
end
end
p nums
解决了 只要不对 nums 重新赋值就好了
# @param {Integer[]} nums
# @param {Integer} k
# @return {void} Do not return anything, modify nums in-place instead.
def rotate(nums, k)
if k!=0
k %=nums.length
t = nums.pop k
t.reverse.each do |value|
nums.insert 0,value
end
end
return
end