[LeetCode]String to Integer (atoi)

Description:
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 a const 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.

解析：个人感觉题目描述的不够清楚，所以一开始可能就没有想的那么全。这道题如果边界值都考虑清楚了，还是比较简单的。也没有用到什么特殊的算法。
当然，如果追求代码的简洁度还是很有挑战的，这里直接取leetcode大神的解法学习下。

Code：

public int myAtoi(String str) {
int index = 0, sign = 1, total = 0;
//1. Empty string
if(str.length() == 0) return 0;
//2. Remove Spaces 这里可以用str=str.trim()取代
while(str.charAt(index) == ' ' && index < str.length())
index ++;
//3. Handle signs
if(str.charAt(index) == '+' || str.charAt(index) == '-'){
sign = str.charAt(index) == '+' ? 1 : -1;
index ++;
}
//4. Convert number and avoid overflow
while(index < str.length()){
int digit = str.charAt(index) - '0';
if(digit < 0 || digit > 9) break;
//check if total will be overflow after 10 times and add digit
if(Integer.MAX_VALUE/10 < total || Integer.MAX_VALUE/10 == total && Integer.MAX_VALUE %10 < digit)
return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
total = 10 * total + digit;
index ++;
}
return total * sign;
}

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