【LeetCode】8. String to Integer (atoi)

1.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 (−231) is returned.

2.分析

读完题目,主要涉及到字符串的简单处理,如:将字符串首位两端空白字符去出,数字符号位取出等以注意实际编程中$int$类型数据的范围,避免出现溢出。其它注意事项,可参照3.编程实现。

3.编程实现

public class StringtoInteger {
    public int myAtoi(String str) {
        if(str == null ||  str.trim().length()==0) return 0;
        str = str.trim();
        char firstChar = str.charAt(0);     // 获取符号字符
        int sign = 1;                       // 正数定义为1
        int start = 0;                      // 数字部分起始索引
        long res = 0;                       // 避免字符串数字越界
        if(firstChar == '+') {
            sign = 1;
            start ++;
        } else if(firstChar == '-') {
            sign = -1; // 负数
            start ++;
        }
        // 开始遍历
        for (int i= start;i < str.length(); i++){
            if(!Character.isDigit(str.charAt(i))){
                return (int)res*sign;
            }
            res = res*10 + str.charAt(i)-'0';
            if (sign == 1 && res > Integer.MAX_VALUE)  return Integer.MAX_VALUE;
            if (sign == -1 && res*-1 < Integer.MIN_VALUE) return Integer.MIN_VALUE;
        }
        return (int)res*sign;
    }
    @Test
    public void test(){
        StringtoInteger sti = new StringtoInteger();
        System.out.println(sti.myAtoi("-91283472332"));
    }
}

4.结果分析

由于其遍历的次数为数字的位数,则其时间复杂度为$O(n)$,在空间复杂度上,空间复杂度为$O(1)$

5.个人订阅号

第一个订阅号侧重于:使用c/c++/java去实现一些经典算法,以及javaEE开发,Linux服务器操作等。第二个订阅号主要侧重于机器学习,深度学习,数学,自然语言处理等方面的知识。







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