1.功能实现
之前文章有，可以点开看看
2.应用(1)括号匹配及后缀表达式
class Solution(object):
def isValid(self, s):
"""
:type s: str
:rtype: bool
"""
stack = []
d = ["()", "[]", "{}"]
for i in xrange(0, len(s)):
stack.append(s[i])
if len(stack) >= 2 and stack[-2]+stack[-1] in d:
stack.pop()
stack.pop()
return len(stack) == 0150. Evaluate Reverse Polish Notation（计算逆波兰表达式RPN）
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
后缀
class Solution:
def evalRPN(self, tokens):
"""
:type tokens: List[str]
:rtype: int
"""
stack = []
for s in tokens:
if s in ['+','-','*','/']:
b = stack.pop()
a = stack.pop()
if s == '+':
stack.append(a+b)
elif s == '-':
stack.append(a-b)
elif s == '*':
stack.append(a*b)
else:
stack.append(int(a/b))
else:
stack.append(int(s))
ans = int(stack.pop())
return ans
中缀
class Solution:
def evalRPN(self, tokens):
"""
:type tokens: List[str]
:rtype: int
"""
stack = []
for s in tokens:
if s in ['+','-','*','/']:
a = stack.pop(0)
b = stack.pop(0)
if s == '+':
stack.append(a+b)
elif s == '-':
stack.append(a-b)
elif s == '*':
stack.append(a*b)
else:
stack.append(int(a/b))
else:
stack.append(int(s))
ans = int(stack.pop())
return ans
3.表达式计算与函数调用栈
Leetcode 224. Basic Calculator （基础计算器）
Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .
You may assume that the given expression is always valid.
Some examples:
"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23
class Solution:
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
stack = []
l = len(s)
i = 0
sgn,res = 1, 0 # 初始化res和符号变量sgn 存储前一个符号
while i <= l-1: # 遍历s
ch = s[i]
if ch ==' ':
i += 1
elif ch.isdigit(): # 处理遇到的数字（只可处理整数，其他的自行修改）
j = i
while j <= l-1:
j += 1
if j==l or not s[j].isdigit():
break
num = int(s[i:j]) # 此时的操作数为num
i = j
elif ch =='+' or ch == '-': # 遇到符号，对前面的结果进行计算
res += sgn * num
sgn = 1 if ch=='+' else -1 # 更新符号变量 sgn
i += 1
num = 0
elif ch == '(': # 左括号，将当前结果和符号 压入栈中
stack.append(res)
stack.append(sgn)
sgn,res = 1, 0 # 重置res和sgn
i += 1
elif ch == ')':
res += sgn*num # 遇到右括号，对前面的结果进行计算
res *= stack.pop() # 并取出栈中存储的res和sgn
res += stack.pop()
i += 1
num = 0 # 注意重置num=0 ,否则右括号后面第一个运算符会重复计算num
if num!= 0: # 如果此时num不为0 继续计算
res += sgn*num
return res
385. Mini Parser （迷你解析器）
Given a nested list of integers represented as a string, implement a parser to deserialize it.
Each element is either an integer, or a list -- whose elements may also be integers or other lists.
Note: You may assume that the string is well-formed:
String is non-empty.
String does not contain white spaces.
String contains only digits 0-9, [, - ,, ].
Example 1:
Given s = "324",You should return a NestedInteger object which contains a single integer 324.
Example 2:
Given s = "[123,[456,[789]]]",Return a NestedInteger object containing a nested list with 2 elements:
1. An integer containing value 123.
2. A nested list containing two elements:
i. An integer containing value 456.
ii. A nested list with one element:
a. An integer containing value 789.
def deserialize(self, s):
return NestedInteger(s) if isinstance(s, int) else reduce(lambda a, x: a.add(self.deserialize(x)) or a, s, NestedInteger()) if isinstance(s, list) else self.deserialize(eval(s))