实现atoi这个函数，将一个字符串转换为整数。如果没有合法的整数，返回0。如果整数超出了32位整数的范围，返回INT_MAX(2147483647)如果是正整数，或者INT_MIN(-2147483648)如果是负整数。
样例
"10" =>10
"-1" => -1 "123123123123123" => 2147483647 "1.0" => 1标签
字符串处理 基本实现 优步
code
class Solution {public: /** * @param str: A string * @return An integer */ int atoi(string str) { // write your code here str = eraseSpace(str); int size=str.size(); if(size <= 0) return 0; if(str[0]<='9' && str[0]>='0') { int num=str[0]-'0', i, dec=str[0]-'0'; for(i=1; i ='0') { dec = num; num = num * 10; if(isCrossBorder(num, dec)) return 2147483647; dec = num; num += str[i]-'0'; if(isCrossBorder(num, dec)) return 2147483647; } else { return num; } } return num; } else if(str[0] == '+') { int num=0, i, dec=0; for(i=1; i ='0') { dec = num; num = num * 10; if(isCrossBorder(num, dec)) return 2147483647; dec = num; num += str[i]-'0'; if(isCrossBorder(num, dec)) return 2147483647; } else { return num; } } return num; } else if(str[0] == '-') { int num=0, i, dec=0; for(i=1; i ='0') { dec = num; num = num * 10; if(isCrossBorder(num, dec)) return -2147483648; dec = num; num += str[i]-'0'; if(isCrossBorder(num, dec)) return -2147483648; } else { return num*-1; } } return num*-1; } else { return 0; } } bool isCrossBorder(int num1, int num2) { if(num1 < num2) return true; return false; } string eraseSpace(string str) { int size=str.size(),i=0,mark=0; string result; for(i=0; i