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【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

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.

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C++ /* * 問題：atoi字符串轉數字函數實現 * 要求：先檢測第一個非空格字符，然後識別“+”或“-”，將後面的數字字符翻譯為數字，終止於非數字字符 * 數值範圍：[?2^31, 2^31 ? 1] 即[INT_MIN, INT_MAX] */ /* * 分析參考： I think we only need to handle four cases: discards all leading whitespaces sign of the number overflow invalid input */ #include<climits> class Solution { public: int myAtoi(string str) { int sign = 1; //符號 long long base = 0; //結果 int i = 0; //用於遍歷字符串 if(str.empty()) return 0; while(str[i] == ‘ ‘) i++; //跳過空格 if(str[i] == ‘+‘ || str[i] == ‘-‘)//翻譯正負號 { sign = (str[i] == ‘-‘)? -1: 1; i++; } while(str[i] >= ‘0‘ && str[i] <= ‘9‘) //數字字符時才進入循環 { int digit = str[i] - ‘0‘; //翻譯數字字符 base = 10*base + digit; //處理可能的溢出情況 if( (sign>0 && base>INT_MAX) || (sign<0 && -base<INT_MIN) ) { return (sign==1)? INT_MAX:INT_MIN; } i++; } return sign*base; } };

【LeetCode & 劍指offer刷題】字符串題7：String to Integer (atoi)