LeetCode--String to Integer (atoi)

String to Integer (atoi)## [更多技术博客 http://vilins.top/](http://vilins.top/)1 题目Implementatoiwhich 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: 42Example 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.2 分析这题将string类型的字符转变为int类型其实转换的过程不复杂只是限制条件太多需要自己找出那些转换是合法的有一下几个部分需要注意的。2.1 特殊的情况1. 形如--2之类的string是不合法的不能做出转换返回0.2. 形如138之类的转换是合法的返回13.3. 形如-2-之类的转换的合法的返回-2.注意上面的几种情况解出这道题目就不难了。2.2数据的溢出在这整个过程中string转化为int类型的数据可能会大于long long 类型所以我们在sum循环相加的时候就应该开始做出判断否则到最后数据溢出。3 源码class Solution { public: int getIndex(string str) { int index -1; for(int i 0; i str.size(); i) { if(str[i] ! ) { index i; break; } } return index; } int getLastIndex(int index, string str) { int lastIndex str.size(); int tag 0; for(int i index; i str.size(); i) { if((str[i] 0||str[i] 9)!(str[i] -||str[i] )) { lastIndex i; break; } } //cout lastIndex lastIndex endl; return lastIndex; } int myAtoi(string str) { int index getIndex(str); //cout index index endl; int flag 0; string target; if(index -1||str[index] ! -str[index] ! (str[index] 0||str[index] 9)) { return 0; } else { int lastIndex getLastIndex(index, str); if(str[index] -) { flag -1; target str.substr(index1, lastIndex-index-1); } else if(str[index] ) { flag 1; target str.substr(index1, lastIndex-index-1); } else { target str.substr(index, lastIndex-index); } int tag 0; for(int i 0; i target.size(); i) { if(target[i] 0||target[i] 9) { lastIndex i; break; } } target target.substr(0, lastIndex); //cout target endl; } long long sum 0; for(int i 0; i target.size(); i) { if(target[i] 0||target[i] 9) { return 0; } sum sum*10 (target[i] - 0); if(sum INT_MAX) { break; } } //cout flag flag endl; if(flag -1) { sum 0-sum; } //cout sum endl; if(sum INT_MIN) { return INT_MIN; } else if(sum INT_MAX) { return INT_MAX; }else { return sum; } } };## [更多技术博客 http://vilins.top/](http://vilins.top/)