Leetcode题解3：字符串转整数

题目

Implement atoi which converts a string to an integer.

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.

Note:

• Only the space character ' ' is considered as whitespace character.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹ − 1]. If the numerical value is out of the range of representable values, INT_MAX (2³¹ − 1) or INT_MIN (−2³¹) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−2^31^) is returned.

题目大意：实现字符串转换成为整数的atoi函数，函数有如下要求：

• 丢弃开头的若干个字符串直到遇到第一个非空格字符串，只有空格才算空格字符
• 第一个非空格字符可以是+、-代表整数的符号
• 整数后面可以跟若干个非整数字符，这些字符被忽略
• 如果第一个非空格字符或者字符串为空，返回0
• 只返回-2³¹到2³¹-1之间的数字，超过的返回-2³¹或者2³¹-1

解题思路

从开头逐个字符处理，如果是空格，则忽略，如果是非数字或者正负号，则返回0，如果是数字或者正负号则将数字或者正负号转换成整数。

class Solution {
    public int myAtoi(String str) {
        boolean sign = true,firstInt=true,firstC=true; // 默认是正数,是不是第一个非空格字符
        int base = 10;
        long ret = 0;
        for(int i = 0;i<str.length();i++){
            char c = str.charAt(i);
            if(firstC&&c == ' ') continue;
            else if(firstC&&c == '+'){ // 不是第一个非空格字符略过这个字符
                firstC = false;
                sign = true;
            }else if(firstC&&c == '-'){
                firstC = false;
                sign = false;
            }else if(c >= '0' && c <= '9'){
                firstC = false;
                if(firstInt&&c=='0') continue; // 去掉开头的0
                firstInt = false; // 如果开头的数字不是0了，置为false
                if(sign){
                    ret = ret*base + (c-'0');
                }else{
                    ret = ret*base - (c-'0');
                }
                if(ret > Integer.MAX_VALUE){ // 在计算的时候就进行判断，避免64位整数也溢出
                    return Integer.MAX_VALUE;
                } 
                if(ret < Integer.MIN_VALUE){
                    return Integer.MIN_VALUE;
                }
            }else {
                if(ret <= Integer.MAX_VALUE && ret >= Integer.MIN_VALUE) return (int)ret;
                else{
                    if(ret > 0) return Integer.MAX_VALUE;
                    else return Integer.MIN_VALUE;
                }
            }
        }
        if(ret <= Integer.MAX_VALUE && ret >= -Integer.MIN_VALUE) return (int)ret;
        else{
            if(ret > 0) return Integer.MAX_VALUE;
            else return Integer.MIN_VALUE;
        }
    }
}