LeetCode 之 JavaScript 解答第8题 —— 字符串转换整数 （String to Integer ）

Time：2019/4/19

Title: String To Integer

Difficulty: Medium

Author: 小鹿

题目：String To Integer（字符串转换整数 (atoi)）

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:

' '

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.

Solve:

▉ 问题分析

▉ 算法思路

1）空格：如果当前的字符为空格，直接跳过，判断下一字符。
2）符号位：如果判断当前的字符为 “ + ” 或 “ - “，用 sign 存储 1 或 -1，最后乘最后得出的数字。
3）只取数字：判断当前是否满足条件（0
  
▉ 测试用例
  
2）带空格、负数的字符串
3）带字母的字符串
4）空字符串
5）超出限制的字符串

  
▉ 代码实现
  
var myAtoi = function(str) {
    // 最大值与最小值
    let MAX_VALUE = Math.pow(2,31)-1;
    let MIN_VALUE = -Math.pow(2,31);
    // 判断字符窗是否为空
    if(str == null || str.length == 0) return 0;
    // 初始化
    // sign：记录正负号
    // base: 记录数字的位数
    let [index,base,sign,len] = [0,0,1,str.length];

    // 跳过空格
    while(index < len && str.charAt(index) == ' '){
        index++;
    }

    // 获取符号位
    if(index < len && (str.charAt(index) == "+") || str.charAt(index) == '-'){       // 记录正负号
        sign = 1 - 2 * ((str.charAt(index++) == '-') ? 1 : 0);
    }

    // 只取数字,碰到非数字退出循环
    while(index < len && parseInt(str.charAt(index)) >= 0 && parseInt(str.charAt(index)) <= 9){
        // 溢出判断，MAX_VALUE 的个位为 7
        if(base > parseInt(MAX_VALUE/10) || (base == parseInt(MAX_VALUE/10) && parseInt(str.charAt(index)) > 7)){
            if(sign == 1){
                return MAX_VALUE;
            }else{
                return MIN_VALUE;
            }
        } 
        // 记录位数
        base = base * 10 + parseInt(str.charAt(index++));                
    }
    // 返回符号位 * 当前字符串中的数字
    return sign * base;
};
  
▉ 考查内容
  
1）对字符串的基本操作。
  
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