输入一个字符串,包括数字字母符号,可以为空
如果是合法的数值表达则返回该数字,否则返回0
+2147483647 1a33
2147483647 0
class Solution {public: int StrToInt(string str) { if(str.empty()){ return 0; } int symbol=1; if(str[0]=='-'){ symbol=-1; str[0]='0'; }else if(str[0]=='+'){ symbol=1; str[0]='0'; } int sum=0; for(int i=0;i<str.size();i++){ if(str[i]<'0'||str[i]>'9'){ sum=0; break; } sum=sum*10+str[i]-'0'; } return symbol*sum; }};
边界条件:数据上下 溢出空字符串只有正负号有无正负号错误标志输出class Solution {public: enum Status{kValid = 0,kInvalid}; int g_nStatus = kValid; int StrToInt(string str) { g_nStatus = kInvalid; long long num = 0; const char* cstr = str.c_str(); if( (cstr != NULL) && (*cstr != '\0') ) { int minus = 1; if(*cstr == '-') { minus = -1; cstr++; } else if(*cstr == '+') cstr++; while(*cstr != '\0') { if(*cstr > '0' && *cstr < '9') { g_nStatus = kValid; num = num*10 + (*cstr -'0'); cstr++; if( ((minus>0) && (num > 0x7FFFFFFF)) || ((minus<0) && (num > 0x80000000)) ) { g_nStatus = kInvalid; num = 0; break; } } else { g_nStatus = kInvalid; num = 0; break; } } if(g_nStatus == kValid) num = num * minus; } return (int)num; }};
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