LeetCode:String to Integer (atoi)

Posted 2020-07-26 walker lee

String to Integer (atoi)

 

Total Accepted: 106573 Total Submissions: 784972 Difficulty: Easy

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.

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思路：

处理好以下四个要素：

1.空串、“0”串处理

2.去除前面空格

3.判断符号

4.字符转换与溢出处理

java code:

public class Solution {
    public int myAtoi(String str) {
        
        int ans = 0;
        int sign = 1;
        int index = 0;
        // 1.空串、“0”串处理
        if(str==null || str.length()==0) return 0;
        int len = str.length();
        
        // 2.去除前面空格
        while(str.charAt(index) == ' ' && index < len) index++;
        
        // 3.判断符号
        if(str.charAt(index) == '-' || str.charAt(index) == '+') {
            sign = str.charAt(index) == '+' ? 1 : -1;
            index++;
        }
        
        // 4.字符转换与溢出处理
        while(index < len) {
            int digit = str.charAt(index) - '0';
            if(digit < 0 || digit > 9) break;
            // 溢出
            while(Integer.MAX_VALUE/10 < ans 
                || (Integer.MAX_VALUE/10 == ans && Integer.MAX_VALUE%10 < digit)) {
                    
                return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            ans = 10 * ans + digit;
            index++;
        }
        return sign * ans;
    }
}