LeetCode (力扣) 8. String to Integer (atoi) ( C ) - Medium

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题目为给定一个字符串，把它转换成整数型态，复现 C语言中的 atoi() 函数。

这题的测资有点复杂，要考虑的情况有点多。

题目与范例如下

 

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

The algorithm for myAtoi(string s) is as follows:

Read in and ignore any leading whitespace.
Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
Return the integer as the final result.
Note:

Only the space character ' ' is considered a whitespace character.
Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.


Example 1:

Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
^
Step 3: "42" ("42" is read in)
^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.
Example 2:

Input: s = " -42"
Output: -42
Explanation:
Step 1: " -42" (leading whitespace is read and ignored)
^
Step 2: " -42" ('-' is read, so the result should be negative)
^
Step 3: " -42" ("42" is read in)
^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.
Example 3:

Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
Example 4:

Input: s = "words and 987"
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.
Example 5:

Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
^
Step 3: "-91283472332" ("91283472332" is read in)
^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.


Constraints:

0 <= s.length <= 200
s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

 

我的解题方法为先考虑溢位的问题，在考虑空白跟正负号，并于判断正负号的函式内，判断后面是否有接正负号或其他字元。

 

下方为我的代码

int myAtoi(char * s){
    int total = 0;
    bool negative = false,yn = false;
    for(int i = 0;i<strlen(s);i++){
        if(s[i]==' '){
            if(yn == true)
                break;
        }
        else if(s[i]=='-'){
            if(yn == true){
                if(s[i-1]=='-'||s[i-1]=='+')
                    return 0;
                else
                    return total;
            }
            negative = true;
            yn = true;
        }
        else if(s[i]=='+'){
            if(yn == true){
                if(s[i-1]=='-'||s[i-1]=='+')
                    return 0;
                else
                    return total;
            }
            yn = true;
        }
        else if(s[i]>='0'&&s[i]<='9'){
            yn = true;
            if(total > 214748364 ||  total < -214748364){
                if(negative==true)
                    return INT_MIN;
                else
                    return INT_MAX;
            }
            else{
                total*=10;
                if(negative == true){
                    if(total<-2147483648+(s[i]-'0')){
                        return INT_MIN;
                    }
                    total-=(s[i]-'0');
                }
                else{
                    if(total>2147483647-(s[i]-'0')){
                        return INT_MAX;
                    }               
                    total += (s[i]-'0');
                }
            }
        }
        else{
            break;
        }
    }
    
    return total;
}

 

下方为时间与空间之消耗

Runtime: 4 ms, faster than 59.13 % of C online submissions for String to Integer ( atoi ) .

Memory Usage: 5.5 MB, less than 92.37 % of C online submissions for String to Integer ( atoi ).