108 lines
2.3 KiB
C
108 lines
2.3 KiB
C
#include <stdio.h>
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#include <string.h>
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#include <stdlib.h>
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#include <math.h>
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/**
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* https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/5/strings/37/
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* 字符串转整数（atoi）
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*/
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#define INT_MIN -2147483648
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#define INT_MAX 0x7fffffff
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int myAtoi(char *str)
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{
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long long ret = 0;
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char flag = 1, len = 0;
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while (*str == ' ') //删除空
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{
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str++;
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}
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if (*str == '-')
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{
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flag = -1;
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str++;
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}
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else if (*str == '+')
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{
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str++;
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}
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else if (*str < '0' || *str > '9')
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{
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return 0;
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}
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while (*str >= '0' && *str <= '9' && len < 11)
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{
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ret = ret * 10 + (*str - '0');
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str++;
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if (ret != 0)
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{
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len++;
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}
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}
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ret *= flag;
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if (ret < INT_MIN)
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{
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return INT_MIN;
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}
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if (ret > INT_MAX)
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{
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return INT_MAX;
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}
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return ret;
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}
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/*
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int myAtoi(char* str) {
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if(strlen(str)==0)
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return 0;
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char* xianzai= str;
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long long fanhui = 0;
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int flag = 1;//记录符号位正负
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long long max = pow(2,31)-1;
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long long min = pow(2,31);//溢出的上下界不相同
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while (*xianzai == ' ')
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{
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xianzai++;//跳过空格位置
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}
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if (*xianzai == '-')
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{
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flag = -1;
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xianzai++;
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}
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else if(*xianzai == '+')
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{
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xianzai++;
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}
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else if (*xianzai < '0' || *xianzai > '9')
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{
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return 0;
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}
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while (*xianzai >= '0' && *xianzai <= '9')
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{
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fanhui= fanhui*10;
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fanhui=fanhui+*xianzai-'0';//利用符号本身的ascii码直接转换成数字
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if(flag==1 && fanhui>max)
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{
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//正数溢出，返回INT_MAX
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return max;
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}
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else if(flag==-1&&fanhui>min)
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{
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return min;//注意int类型上下界数值上相差1
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}
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xianzai++;
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}
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return flag*fanhui;
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}
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*/
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int main()
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{
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printf("%d\n", myAtoi("-1"));
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printf("%d\n", myAtoi("456"));
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printf("%d\n", myAtoi(" -42"));
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printf("%d\n", myAtoi("a65r"));
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printf("%d\n", myAtoi("9223372036854775808"));
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printf("%d\n", myAtoi("-91283472332"));
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printf("%d\n", myAtoi("4193 with words"));
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printf("%ld,%ld\n", -pow(2, 31), pow(2, 31) - 1);
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getchar();
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return 0;
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} |