# 8. String to Integer (atoi)

# Tags

1. String
2. Math

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.

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.

# 题意：

实现atoi函数，将字符串转换为整数。

# 分析：

题目很简单，就一句话，哪什么叫atoi呢？我搜索了一下，发现它是alphanumeric to integer的简称，翻译过来就是把字符串转换成整型数的意思，也就是说它代表的就是这么一个函数，那么我们需要做的就是将字符串转换成一个整数，那么涉及到的第一个问题就是数字的边界问题了，JavaScript中Number的数值范围是-1.7976931348623157E+308~1.7976931348623157E+308之间,所以不可能这么大，还好要求中明确的限定了最大最小数的范围。 第二问题就是整数是有正有负的，所以需要判断正负情况。

# 思路：

根据上面的分析下来，思路其实已经比较清晰了：

1. 判断字符串是否有正负符号，然后定义一个变量做标记，最后根据标记再给求出的数字转换正负；
2. 设定要求中的最大最小值边界，如果越界了，要做限定；
3. 字符串都是左往右的，所以转成数字大小的时候切记反过来，字符的低位就是数字的高位。而且每位字符转成数字的条件必须满足在0~9之间

# Js实现

# 复杂度

时间复杂度O(n)

/**
 * @param {string} str
 * @return {number}
 */
let myAtoi = function(str) {

  let INT_MAX = 2147483647;
  let INT_MIN = -2147483648;
  let sign = 1,
    index = 0,
    num = 0;
  //去除空格
  str = str.trim();
  if (str.length === 0 || str === null) {
    return 0;
  }

  if (str.charAt(index) == '+') {
    index++;
  } else if (str.charAt(index) == '-') {
    sign = -1;
    index++;
  }

  while (str.length > index && str.charAt(index) >= '0' && str.charAt(index) <= '9') {
    num = num * 10 + (str.charAt(index) - '0');
    index++;
  }

  if (sign === -1) {
    num = -num;
  }
  if (num > INT_MAX) {
    return INT_MAX;
  }
  if (num < INT_MIN) {
    return INT_MIN;
  }

  return num;
};