0%

[Leetcode]8. String to Integer (atoi)(C++)

题目描述

题目链接: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.

例子

例子 1

Input: str = “42”
Output: 42

例子 2

Input: str = “ -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.

例子 3

Input: str = “4193 with words”
Output: 4193
Explanation: Conversion stops at digit ‘3’ as the next character is not a numerical digit.

例子 4

Input: str = “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.

例子 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.

Follow Up

Note

  • Only the space character ' ' is considered a whitespace character.
  • Assume we are dealing with an environment that 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.

Constraints

  • 0 <= s.length <= 200
  • s consists of English letters (lower-case and upper-case), digits, ' ', '+', '-' and '.'.

解题思路

这道题由于只需要处理整数部分,而且结束条件也很清楚,所以相对来说比较简单。通过指针遍历数组,首先略过所有前面的空格。当遇到第一个非空格字符时判断是不是正负号或者数字,如果不是直接返回结果。加入是正负号则通过一个布尔变量存储,接下来继续遍历所有剩下字符直到遇到不是数字的字符为止。边遍历边更新结果,这里注意一下由于结果可能会超过 int 的范围,所以先用 long 存储结果。同时在便利过程中检查是否超限,如果超限可以提前返回结果。结束遍历之后根据正负号返回相应结果即可,代码如下:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
#include <string>

class Solution {
public:
int myAtoi(std::string s) {
long result = 0;
int ptr = 0;
while (ptr < s.length() && s[ptr] == ' ') ptr++;
if (ptr == s.length() || (s[ptr] != '+' && s[ptr] != '-' && !std::isdigit(s[ptr]))) return result;
bool is_negative = false;
if (s[ptr] == '+' || s[ptr] == '-') {
is_negative = s[ptr] == '-';
ptr++;
}

while (ptr < s.length()) {
if (!isdigit(s[ptr])) break;

result *= 10;
result += s[ptr] - '0';

ptr++;

if (!is_negative && result > INT32_MAX) {
return INT32_MAX;
}
if (is_negative && -result < INT32_MIN) {
return INT32_MIN;
}
}
if (is_negative) result = -result;
return result;

}
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 8.StringToIntegerAtoi.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions