[leetcode题解]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.

就是字符串直接转整数,我们可以使用for循环的方式做,也可以使用类似状态机的方式做,因为这里其实状态不多,我们不单独列出所有状态,直接判断进行状态判决。

int myAtoi(char* str) 
{
	int len = strlen(str);
	if (len == 0)return 0;

	int sign = 1;
	int i = 0;
	while (str[i] == ' ')++i;
	if (i == len)return 0;
	if (str[i] == '-')
	{
		sign = -1;
		++i;
	}
	else if (str[i] == '+')++i;

	long long res = 0;
	while (i < len && '0' <= str[i] && str[i] <= '9')
	{
		res *= 10;
		res += sign * (str[i] - '0');
		if (res > INT_MAX)
		{
			res = INT_MAX;
			break;
		}
		else if (res < INT_MIN)
		{
			res = INT_MIN;
			break;
		}
		++i;
	}
	return res;
}

类似状态机的方式解析字符串:

struct context
{
    char *s;
    int len;
    int sign;
    long long num;
    bool valid;
};

void white_skip(struct context *ctx) {
    while (ctx->len >= 0) {
        if (*ctx->s == ' '
           || *ctx->s == '\t') {
            ctx->s++;
            ctx->len--;
        } else {
            break;
        }
    }
}

void parse_sign(struct context *ctx) {
    if (ctx->len <= 0)
        ctx->sign = 1;
    
    if (*ctx->s == '-') {
        ctx->s++;
        ctx->len--;
        ctx->sign = -1;
    } else if (*ctx->s == '+') {
        ctx->s++;
        ctx->len--;
        ctx->sign = 1;
    }
}

void parse_value(struct context *ctx) {
    long long num = 0;
    
    while (ctx->len >= 0) {
        char c = *ctx->s;
        if (c < '0' || c > '9')
            break;
        ctx->num *= 10;
        ctx->num += (c - '0');
        ctx->valid = true;
        ctx->s++;
        ctx->len--;
        num = ctx->num * ctx->sign;
        if (num > INT_MAX || num < INT_MIN)
            break;
            
    }
    if (ctx->valid) {
        num = ctx->num * ctx->sign;
        if (num > INT_MAX)
            ctx->num = INT_MAX;
        else if (num < INT_MIN)
            ctx->num = INT_MIN;
        else
            ctx->num = num;
    }
}

int myAtoi(char* str) {
    struct context ctx;
    ctx.s = str;
    ctx.len = strlen(str);
    ctx.sign = 1;
    ctx.num = 0;
    ctx.valid = false;
    
    white_skip(&ctx);
    parse_sign(&ctx);
    parse_value(&ctx);
    
    if (!ctx.valid)
        return 0;
    
    return ctx.num;
}

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本文标题:[leetcode题解]8. String to Integer (atoi)

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