8. String to Integer (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:

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.

Solution:

思路:

1, 除去' ', 之前是使用whil循环判断, 可直接用str.find_first_not_of(' ')找到第一个非空字符位置

2, 用一个flag标记正负, 如果没有+-号, flag缺省为1

3, ans等于扩大10倍+当前位, 如果溢出就要返回INT_MIN或者INT_MAX

错误的几个case

/*
// case1
Input:
"+-2"
Output:
2
Expected:
0
*/


/*
// case2
Input:
"2147483648"
Output:
-2147483648
Expected:
2147483647
*/

/*
// case3
Input:
"-91283472332"
Output:
-1089159116
Expected:
-2147483648
*/

最后的代码:

// Time:O(n), Space:O(1)
class Solution {
 public:
  int myAtoi(string str) {
    long long int ans = 0;                        // case3
    int flag = 1;
    int i = str.find_first_not_of(' ');
    if (str[i] == '-' || str[i] == '+') {         // case1
      flag = (str[i++] == '-') ? -1 : 1;
    }
    while ('0' <= str[i] && str[i] <= '9') {
      ans = ans*10 + (str[i++] - '0');
      if (ans * flag >= INT_MAX) return INT_MAX;  // case2
      if (ans * flag <= INT_MIN) return INT_MIN;
    }
    return ans * flag;
  }
};