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.
- 【函数说明】atoi() 函数会扫描 str 字符串，跳过前面的空白字符（例如空格，tab缩进等），直到遇上数字或正负符号才开始做转换，而再遇到非数字或字符串结束时(’\0’)才结束转换，并将结果返回。
- 【返回值】返回转换后的整型数；如果 str 不能转换成 int 或者 str 为空字符串，那么将返回 0。如果超出Integer的范围，将会返回Integer最大值或者最小值。
- 【处理思路】按照函数说明来一步步处理。首先判断输入是否为null。然后使用trim()函数删掉空格。判断是否有正负号，做一个标记。返回的是整形数，可以使用double来暂存结果。按位来计算出结果。如果遇到非数字字符，则返回当前结果。加上前面的正负号。结果若超出了整形范围，则返回最大或最小值。最后返回处理结果。
class Solution {
public int myAtoi(String str) {
str = str.trim(); // kill add white spaces
if (str == null || str.length() < 1) {
return 0;
}
int i = 0; // index of str
char flag = '+'; // default positive
if (str.charAt(0) == '-') {
flag = '-';
i++;
} else if (str.charAt(0) == '+') {
i++;
}
double res = 0;
// abandon the non-digit char; calculate the result
while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
res = res * 10 + str.charAt(i) - '0';
i++;
}
if (flag == '-') res = -1 * res;
if (res > Integer.MAX_VALUE) {
return Integer.MAX_VALUE;
} else if (res < Integer.MIN_VALUE) {
return Integer.MIN_VALUE;
}
return (int) res;
}
}