LeetCode——String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

spoilers alert... click to show requirements for atoi.

Requirements for atoi:

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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

实现 atoi 将字符串转为整数。

需要考虑正负号，空格，非数字字符出现的话，其后部分随之忽略。

提交了如下的代码，结果运行超时：

	public static int atoi(String str) {
		str = str.trim();
		if(str.isEmpty())
			return 0;
		char[] ch = str.toCharArray();
		for(int i=1;i<ch.length;i++){
			if(!Character.isDigit(ch[i])){
				str = str.substring(0, i);
				break;
			}
		}
		ch = str.toCharArray();
		int ret = 0;
		for (int i = 0; i < ch.length; i++) {
			if (ch[0] == '-') {
				ret = ret * 10 - (ch[i + 1] - 48);
				if (i == ch.length - 2)
					break;
			} else if (ch[0] == '+') {
				ret = ret * 10 + (ch[i + 1] - 48);
				if (i == ch.length - 2)
					break;
			} else {
				ret = ret * 10 + (ch[i] - 48);
			}
		}
		return ret;
	}

下面是抄了网友http://my.oschina.net/jdflyfly/blog/283516的代码来学习一下：

	public static int atoi(String str) {
		int max = Integer.MAX_VALUE;
		int min = -Integer.MIN_VALUE;
		long result = 0;
		str = str.trim();
		int len = str.length();
		if (len < 1)
			return 0;
		int start = 0;
		boolean neg = false;

		if (str.charAt(start) == '-' || str.charAt(start) == '+') {
			if (str.charAt(start) == '-')
				neg = true;
			start++;
		}

		for (int i = start; i < len; i++) {
			char ch = str.charAt(i);

			if (ch < '0' || ch > '9')
				break;
			result = 10 * result + (ch - '0');
			if (!neg && result > max)
				return max;
			if (neg && -result < min)
				return min;

		}
		if (neg)
			result = -result;

		return (int) result;
	}

LeetCode——String to Integer (atoi)