leetcode 8 String to Integer (atoi)
leetcode 8 String to Integer (atoi)
String to Integer (atoi)Total Accepted:52232 Total Submissions:401038 My Submissions
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts aconst char * argument, please click the reload button to reset your code definition.
spoilers alert... click to show requirements for atoi.
Requirements for atoi:
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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
c++ 解决方案：
class Solution {
public:
//consider a case: " +-++--3"
int myAtoi(string str)
{
long result = 0;
int indicator = 1;
for(int i = 0; i<str.size();)
{
i = str.find_first_not_of(' ');
if(str[i] == '-' || str[i] == '+')
indicator = (str[i++] == '-')? -1 : 1;
while('0'<= str[i] && str[i] <= '9')
{
result = result*10 + (str[i++]-'0');
if(result*indicator >= INT_MAX) return INT_MAX;
if(result*indicator <= INT_MIN) return INT_MIN;
}
return result*indicator;
}
}
};python解决方案：
class Solution:
# @return an integer
def atoi(self, str):
string = str
buf = ''
# our list
dg_list = ['0','1','2','3','4','5','6','7','8','9']
dg_signal = ['-','+']
signal = ''
#there is no +/- and 0-9 at first
no_signal = True
no_dig = True
#strip whitespace
for i in string.strip():
#if i in dg_signal judge:
#1:it is the first -/+ and signal = -/+, then set no_signal = False . Start to next i
#2:it it the second -/+ So it is wrong,we return 0
if i in dg_signal:
if no_signal is False:
return 0
if no_signal:
signal += i
no_signal=False
continue
#if i is 0-9, we save it to buf
if i in dg_list:
buf+=i
#but if there is a -/+, and the next char is not dig, eg:+a here we break.
elif no_signal is False:
break
else:
#if it is not in dg_list and dg_signal,and it is also has something eg:+322a99 when i = a and buf = +322. We jsut break a and continue
#add 99 to buf
if len(buf)>0:
break
#if len(buf)<0. eg: -a. We return as the sys tips
if len(buf)<=0:
return 0
#add +/-
if len(buf)>0:
buf=signal+buf
if len(buf)<=0:
return 0
f_result = int(buf)
if f_result >2147483647:
return 2147483647
if f_result<-2147483648:
return -2147483648
return f_resultpython解决方案2：
class Solution:
# @return an integer
def atoi(self, str):
str = str.strip()
str = re.findall('(^[\+\-0]*\d+)\D*', str)
try:
result = int(''.join(str))
MAX_INT = 2147483647
MIN_INT = -2147483648
if result > MAX_INT > 0:
return MAX_INT
elif result < MIN_INT < 0:
return MIN_INT
else:
return result
except:
return 0python解决方案3：
class Solution:
# @return an integer
def atoi(self, str):
str = str.strip()
if len(str) == 0: return 0
r, i, l, s = 0, 0, 0, ''
if str[0] in '+-':
s = str[0]
i = 1
for i in xrange(i, len(str)):
if '0' <= str[i] <= '9':
r = r*10 + ord(str[i]) - ord('0')
l += 1
else:
break
if r == 0 and (s or l == 0):
return 0
elif r > 0 and s == '-':
r *= -1
if r > 2147483647:
r = 2147483647
if r < -2147483648:
r = -2147483648
return r