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Merge pull request youngyangyang04#2862 from yyccPhil/leetcode-150

fix: 150题更正Python解法中使用eval()的方法

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- 0150.逆波兰表达式求值.md

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`‎problems/0150.逆波兰表达式求值.md‎`

Lines changed: 10 additions & 23 deletions

Original file line number	Diff line number	Diff line change
`@@ -188,34 +188,21 @@ class Solution(object):`
`188`	`188`	`return stack.pop()`
`189`	`189`	```
`190`	`190`
`191`		`-另一种可行,但因为使用eval相对较慢的方法:`
	`191`	`+另一种可行,但因为使用eval()相对较慢的方法:`
`192`	`192`	```python
`193`		`-from operator import add, sub, mul`
`194`		`-`
`195`		`-def div(x, y):`
`196`		`- # 使用整数除法的向零取整方式`
`197`		`- return int(x / y) if x * y > 0 else -(abs(x) // abs(y))`
`198`		`-`
`199`	`193`	`class Solution(object):`
`200`		`- op_map = {'+': add, '-': sub, '*': mul, '/': div}`
`201`		`-`
`202`		`- def evalRPN(self, tokens):`
`203`		`- """`
`204`		`- :type tokens: List[str]`
`205`		`- :rtype: int`
`206`		`- """`
	`194`	`+ def evalRPN(self, tokens: List[str]) -> int:`
`207`	`195`	`stack = []`
`208`	`196`	`for token in tokens:`
`209`		`- if token in self.op_map:`
`210`		`- op1 = stack.pop()`
`211`		`- op2 = stack.pop()`
`212`		`- operation = self.op_map[token]`
`213`		`- stack.append(operation(op2, op1))`
	`197`	`+ # 判断是否为数字,因为isdigit()不识别负数,故需要排除第一位的符号`
	`198`	`+ if token.isdigit() or (len(token)>1 and token[1].isdigit()):`
	`199`	`+ stack.append(token)`
`214`	`200`	`else:`
`215`		`- stack.append(int(token))`
`216`		`- return stack.pop()`
`217`		`-`
`218`		`-`
	`201`	`+ op2 = stack.pop()`
	`202`	`+ op1 = stack.pop()`
	`203`	`+ # 由题意"The division always truncates toward zero",所以使用int()可以天然取整`
	`204`	`+ stack.append(str(int(eval(op1 + token + op2))))`
	`205`	`+ return int(stack.pop())`
`219`	`206`	```
`220`	`207`
`221`	`208`	`### Go:`

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