[leetcode] 8. String to Integer (atoi)
2016-07-31 07:41
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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.
解法一:
我考虑的是:(1)empty input (2) 正负数;(3) 是否在int上溢出。
其实还应该考虑: (4) 开头是否有空 (5) 如果遇到一个非数字字符,返回当前结果。。。
另外这里在判断结果是否在int上溢出的时候,要判断res > int_max/10,而不是res*10以后再判断res > int_max。理由很简单,就是res = res*10 + ...以后,即使溢出也会在int区间上取得一个值。
class Solution {
public:
int myAtoi(string str) {
int n = str.size();
if (n==0) return 0;
int res = 0;
int i = 0;
bool isNegative = false;
if (str[0]=='-') {
isNegative = true;
i++;
}
while(i<n){
if (res>INT_MAX/10) return isNegative?INT_MIN:INT_MAX;
res = (res*10) + (str[i]-'0');
4000
i++;
}
return isNegative?-res:res;
}
};
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.
解法一:
我考虑的是:(1)empty input (2) 正负数;(3) 是否在int上溢出。
其实还应该考虑: (4) 开头是否有空 (5) 如果遇到一个非数字字符,返回当前结果。。。
另外这里在判断结果是否在int上溢出的时候,要判断res > int_max/10,而不是res*10以后再判断res > int_max。理由很简单,就是res = res*10 + ...以后,即使溢出也会在int区间上取得一个值。
class Solution {
public:
int myAtoi(string str) {
int n = str.size();
if (n==0) return 0;
int res = 0;
int i = 0;
bool isNegative = false;
if (str[0]=='-') {
isNegative = true;
i++;
}
while(i<n){
if (res>INT_MAX/10) return isNegative?INT_MIN:INT_MAX;
res = (res*10) + (str[i]-'0');
4000
i++;
}
return isNegative?-res:res;
}
};
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