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Title Description
Atoi you to implement a function, it can convert a string to an integer.
First, the function will begin with a space character discard useless if necessary, until the find to the first non-space character so far.
When we find the first non-space character is a positive or negative number, the combination of the symbols as much as possible with consecutive numbers up later, as the sign of integer; if the first non-space character is figures, which directly after the continuous numeric characters are combined to form an integer.
In addition to the string after a valid integer part may also exist extra characters, these characters can be ignored, they should not affect a function.
Note: if the character string in the first non-space character is not a valid integer character string is empty or contains only white space character string, then you will not need to be a function of conversion.
In any case, if the function can not effectively convert, 0 is returned.
Description:
We assume that the environment can store 32-bit signed integer size, then its value range is $ [- 2 ^ {31}, {31} ^ 2 --1] $. If the value exceeds this range, qing return INT_MAX ($ 2 ^ {31} - 1 $) or INT_MIN ($ -2 ^ {31} $).
Example 1
输入: "42"
输出: 42
示例 2:Example 2
输入: " -42"
输出: -42
解释: 第一个非空白字符为 '-', 它是一个负号。
我们尽可能将负号与后面所有连续出现的数字组合起来，最后得到 -42 。Example 3
输入: "4193 with words"
输出: 4193
解释: 转换截止于数字 '3' ，因为它的下一个字符不为数字。Example 4
输入: "words and 987"
输出: 0
解释: 第一个非空字符是 'w', 但它不是数字或正、负号。
因此无法执行有效的转换。Example 5
输入: "-91283472332"
输出: -2147483648
解释: 数字 "-91283472332" 超过 32 位有符号整数范围。
因此返回 INT_MIN (−231) 。answer
public int myAtoi(String str) {
str = str.trim(); // 删除字符串头尾空格
if (str.length() == 0) return 0;
int flag = 1; // 符号位标识
int rev = 0; // 数值（无符号）
int edge = Integer.MAX_VALUE / 10; // 判断数值是否超过范围的边界线，这样写可以节省时间
if (str.charAt(0) == '-') {
flag = -1;
str = str.substring(1, str.length()); // 跳过符号位，可不写第二参数
} else if (str.charAt(0) == '+') {
str = str.substring(1, str.length()); // 跳过符号位，可不写第二参数
} else if (!(str.charAt(0) >= '0' && str.charAt(0) <= '9')) { // 如果开始非空字符不为符号或数字，则直接返回 0
return 0;
}
for (char s : str.toCharArray()) {
if (s >= '0' && s <= '9') {
int n = s - '0'; // 计算字符代表值
if (rev >= edge) { // 超过边界情况较少，故该判断写于外侧
if (flag == 1) {
if (rev > edge || n > 7) return Integer.MAX_VALUE;
} else {
if (rev > edge || n > 8) return Integer.MIN_VALUE;
}
}
rev = rev * 10 + n;
} else {
break;
}
}
return rev * flag;
}Complexity Analysis
- Time complexity: $ O (n) $.
- Space complexity: $ O (n) $.
Notes
The problem is not difficult, note range.