LeetCode:8. String to Integer (atoi)

8. String to Integer (atoi)

• 题目
  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: [−2^31, 2^31 − 1]. If the numerical value is out of the range of representable values, INT_MAX (2^31 − 1) or INT_MIN (−2^31) is returned.

  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.

  Input: “-91283472332”
  Output: -2147483648
  Explanation: The number “-91283472332” is out of the range of a 32-bit signed integer.Thefore INT_MIN (−2^31) is returned.

• 解题思路

  • 本题就是相当于重写atoi这个函数，将字符串转化成整数
  • 本题需要注意的点
    • 将字符串前面的空白去掉
    • 数字的正负
    • 数字溢出
    • 输入无效

• 实现代码

	int myAtoi(string str) {
	    int ans = 0;		// 返回的数字
	    int sign = 1; 		// 返回数字的符号
	    int i = 0;
	    while(str[i] == ' ') { ++i; }
	    if(i < str.length && (str[i] == '+' || str[i] == '-')) {
	    	sign = str[i] == '+' ? 1 : -1;
	    	++i;
	    }
	    while(str[i] >= '0' && str[i] <= '9' && i < str.length()) {
	    	if(ans > INT_MAX / 10 || (ans == INT_MAX / 10  && str[i] > '7'))
	    		return sign == 1 ? INT_MAX : INT_MIN;
	    	ans = ans * 10 + str[i] - '0';
	    	++i;
	    }
		return ans * sign;
	}