描述
用一个数组实现三个栈。你可以假设这三个栈都一样大并且足够大。你不需要担心如果一个栈满了之后怎么办。
您在真实的面试中是否遇到过这个题？ 是
样例
ThreeStacks(5) // create 3 stacks with size 5 in single array. stack index from 0 to 2
push(0, 10) // push 10 in the 1st stack.
push(0, 11)
push(1, 20) // push 20 in the 2nd stack.
push(1, 21)
pop(0) // return 11
pop(1) // return 21
peek(1) // return 20
push(2, 30) // push 30 in the 3rd stack.
pop(2) // return 30
isEmpty(2) // return true
isEmpty(0) // return false我的解法：
比较复杂，将一个数组分成了三份，需要考虑很多判断
class ThreeStacks:
"""
@param: size: An integer
"""
def __init__(self, size):
# do intialization if necessary
self.num = [None] * (3*size)
self.one = self.num[0:size]
self.two = self.num[size:2*size]
self.three = self.num[2*size:3*size]
"""
@param: stackNum: An integer
@param: value: An integer
@return: nothing
"""
def push(self, stackNum, value):
# Push value into stackNum stack
if stackNum == 0:
if self.one[0] is None:
self.one.pop(0)
self.one.append(value)
elif stackNum == 1:
if self.two[0] is None:
self.two.pop(0)
self.two.append(value)
elif stackNum == 2:
if self.three[0] is None:
self.three.pop(0)
self.three.append(value)
"""
@param: stackNum: An integer
@return: the top element
"""
def pop(self, stackNum):
# Pop and return the top element from stackNum stack
if stackNum == 0:
if not self.one:
return []
return self.one.pop()
elif stackNum == 1:
if not self.two:
return []
return self.two.pop()
elif stackNum == 2:
if not self.three:
return []
return self.three.pop()
"""
@param: stackNum: An integer
@return: the top element
"""
def peek(self, stackNum):
# Return the top element
if stackNum == 0:
if not self.one:
return []
return self.one[-1]
elif stackNum == 1:
if not self.two:
return []
return self.two[-1]
elif stackNum == 2:
if not self.three:
return []
return self.three[-1]
"""
@param: stackNum: An integer
@return: true if the stack is empty else false
"""
def isEmpty(self, stackNum):
# write your code here
if stackNum == 0:
return self.peek(stackNum) is None
elif stackNum == 1:
return self.peek(stackNum) is None
elif stackNum == 2:
return self.peek(stackNum) is None
排名第一的算法，比较简洁，在数组内又定义了三个数组
class ThreeStacks:
"""
@param: size: An integer
"""
def __init__(self, size):
# do intialization if necessary
self.size = size
self.stack = [[], [], []]
"""
@param: stackNum: An integer
@param: value: An integer
@return: nothing
"""
def push(self, stackNum, value):
# Push value into stackNum stack
self.stack[stackNum].append(value)
"""
@param: stackNum: An integer
@return: the top element
"""
def pop(self, stackNum):
# Pop and return the top element from stackNum stack
return self.stack[stackNum].pop()
"""
@param: stackNum: An integer
@return: the top element
"""
def peek(self, stackNum):
# Return the top element
return self.stack[stackNum][-1]
"""
@param: stackNum: An integer
@return: true if the stack is empty else false
"""
def isEmpty(self, stackNum):
# write your code here
return len(self.stack[stackNum]) == 0