string to integer (atoi)
Implement atoi which converts a string to an integer.
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned.
Note:
Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231, 231 ? 1]. If the numerical value is out of the range of representable values, INT_MAX (231 ? 1) or INT_MIN (?231) is returned.
Example 1:
Input: "42"
Output: 42
Example 2:
Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
Thefore INT_MIN (?231) is returned.
先來一個簡單的：
class Solution:
def myAtoi(self, s):
"""
:type str: s
:rtype: int
"""
l, sign, op, tmp= list(s), 1, ['-','+'], []
while l and l[0] == ' ':
l.pop(0)
if l and l[0] in op:
sign = op.index(l[0])*2 - 1
l.pop(0)
while l and l[0].isdigit() == True:
tmp.append(l[0])
l.pop(0)
if tmp: return min(max(sign*int(''.join(tmp)),-2**31),2**31-1)
else: return 0
但是這題只需要考慮數(shù)字和符號的情況：
若字符串開頭是空格，則跳過所有空格，到第一個非空格字符，如果沒有，則返回0.
若第一個非空格字符是符號+/-，則標記sign的真假，這道題還有個局限性，那就是在c++里面，+-1和-+1都是認可的，都是-1，而在此題里，則會返回0.
若下一個字符不是數(shù)字，則返回0. 完全不考慮小數(shù)點和自然數(shù)的情況，不過這樣也好，起碼省事了不少。
如果下一個字符是數(shù)字，則轉為整形存下來，若接下來再有非數(shù)字出現(xiàn)，則返回目前的結果。
還需要考慮邊界問題，如果超過了整形數(shù)的范圍，則用邊界值替代當前值。
下面的這個方法，解決不了："0-1" 這個的結果是-1。。。
class Solution:
def myAtoi(self, s):
"""
:type str: str
:rtype: int
"""
base = "0123456789"
plus = "+"
minus = "-"
sum = 0
flag = 1
bit = 0
INT_MAX = 2147483647
INT_MIN = -2147483648
if not str:
return 0
if len(str) == 0:
return 0
for letter in str.strip():
if letter in plus:
if bit == 0:
bit = 1
continue
else:
break
elif letter in minus:
if bit == 0:
bit = 1
flag = -1
continue
else:
break
elif letter not in base:
break;
else:
sum *= 10
sum += int(letter)
sum *= flag
if(sum > INT_MAX):
return INT_MAX
if(sum < INT_MIN):
return INT_MIN
return sum
我的方法，有點亂，但是過了全部的test case：
class Solution:
def myAtoi(self, s: str) -> int:
from string import digits
res = list()
s = s.strip()
flag = 1
if len(s) > 0:
if s[0] == '-':
flag = -1
s = s[1:]
elif s[0] == '+':
flag = 1
s = s[1:]
if s == '' or s[0] not in digits:
return 0
for x in s:
if x in digits:
res.append(x)
else:
break
res = int(''.join(res)) * flag
if flag == 1:
if res > (1<<31) -1:
return (1<<31) -1
else:
return res
else:
if res < -(1<<31):
return -(1<<31)
else:
return res
