Leetcode8 字符串转整型

String to Integer(atoi)

Description

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.

• 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: 0Explanation: 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.

Solution

考虑以下四种情况
1)空格
2) 正负号
3)数字
4)上限和下限

Code

class Solution{
 public int myAtoi(String str) {
        int sign = 1, i = 0, r = 0;
        str = str.trim();
        if (str.isEmpty()) return 0;
        else if (str.charAt(i) == '-') { i++; sign = -1; }
        else if (str.charAt(i) == '+') { i++; }
        while (i < str.length() && Character.isDigit(str.charAt(i))) {
            int d = str.charAt(i) - '0';
            if (r > (Integer.MAX_VALUE - d) / 10) return sign > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            r = r * 10 + d;
            i++;
        }
        return r * sign;
    }
}

Appendix

转自 LeetCode Contributor