LeetCode String to Integer (atoi)

来源：互联网发布：游戏制作软件大全编辑：程序博客网时间：2024/05/16 12:09

剑指offer倒数第二道题，用long long 来存整数才能判断相乘是否溢出

#include "stdafx.h"#include <iostream>#include <string>using namespace std;int atoi(const char *str) {int i,signal=1;long long sum=0;if (str==NULL||*str==0){return 0;}while(*str==' '||*str=='\t'||*str=='\n'){str++;}if (*str=='-'||*str=='+'){if (*str=='-'){signal=-1;}str++;}if (*str>'9'||*str<'0'){return 0;}while(*str){if (*str>'9'||*str<'0'){return sum;}if (signal<0){if (10*sum < INT_MIN){return INT_MIN;}sum = sum*10 - *str + '0';//sum*10 + -( *str - '0')if (sum < INT_MIN){return INT_MIN;//-2147483648;-1;}}else{if ( 10 * sum > INT_MAX)//if (10*sum<0){return INT_MAX;}sum = sum*10 + *str - '0';//sum*10 + ( *str - '0')if (sum > INT_MAX){return INT_MAX;//(2147483647);}}str++;}return sum;}int _tmain(int argc, _TCHAR* argv[]){char s[100]="    10522545459";cout<<atoi(s)<<endl;system("pause");return 0;}

以后把主函数也贴上，便于测试了