LeetCode String to Integer (atoi)
class Solution { public: int atoi(const char *str) { if (str == NULL) return 0; long val =0, pos = 0; char ch = '\0'; int stage = 0; // 0-initial, 1-sign-collected, 2-digit-collected, 3-end bool positive = true; int pos_threshold = INT_MAX / 10; int neg_threshold = INT_MIN / 10; while(stage < 3 && (ch = str[pos]) != '\0') { switch(stage) { case 0: // initial stage if (ch == '-' || ch == '+') { // sign detected positive = ch == '+'; stage = 1; pos++; } else if (ch >= '0' && ch <= '9') { // digit detected stage = 2; } else if (ch == ' ' || ch == '\t' || ch == '\n') { // leading white space pos++; } else { // other chars, invalid str to convert stage = 3; } ;break; case 1: // sign-collected stage if (ch >= '0' && ch <= '9') { // digit after sign stage = 2; } else { // other chars, invalid str to convert stage = 3; } ;break; case 2: // number detected stage if (ch >= '0' && ch <= '9') { // digits int pre_val = val; if (positive) { val = val * 10 + (ch - '0'); if (pre_val > pos_threshold || val < pre_val) { stage = 3; val = INT_MAX; } } else { val = val * 10 - (ch - '0'); if (pre_val < neg_threshold || val > pre_val) { stage = 3; val = INT_MIN; } } pos++; } else { // other chars, invalid stage = 3; } ;break; } } return val; } };
没有使用大范围的数据类型，溢出需要考虑好，是个小状态机
第二轮：
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
spoilers alert... click to show requirements for atoi.
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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
看到这种题就有些烦，不过还是按老样子根据各个状态的跳转写起来也还行，代码基本一致
1 class Solution { 2 public: 3 int atoi(string str) { 4 int len = str.size(); 5 if (len == 0) { 6 return 0; 7 } 8 int val = 0; 9 int pre_val = 0; 10 11 int pos_threshold = INT_MAX/10; 12 int neg_threshold = INT_MIN/10; 13 bool positive = true; 14 15 int pos = 0; 16 char ch = 0; 17 // 0-inital, 1-sign read, 2-number read, 3-end 18 int stage = 0; 19 while (pos < len && stage < 3) { 20 ch = str[pos]; 21 switch(stage) { 22 case 0: // inital stage 23 if (ch == '-' || ch == '+') { 24 // sign read 25 stage = 1; 26 positive = ch == '+'; 27 pos++; 28 continue; 29 } 30 if (ch == ' ' || ch == '\t') { 31 // leading white space 32 pos++; //skip them 33 continue; 34 } 35 if (ch >= '0' && ch <= '9') { 36 // number read 37 stage = 2; 38 continue; 39 } 40 //other charaters, invalid case 41 stage = 3; 42 break; 43 44 case 1: // sign read stage 45 if (ch >= '0' || ch <= '9') { 46 // number read 47 stage = 2; 48 continue; 49 } 50 // other character, invalid case(if space between sign and digit is not permited) 51 stage = 3; 52 break; 53 54 case 2: // number read case 55 if (!(ch >= '0' && ch <= '9')) { 56 // character is not digit invalid case 57 stage = 3; 58 continue; 59 } 60 pre_val = val; 61 val = val * 10 + (ch - '0') * (positive ? 1:-1); 62 if (positive) { 63 if (pre_val > pos_threshold || val < pre_val) { 64 // overflow 65 val = INT_MAX; 66 stage = 3; 67 continue; 68 } 69 } else { 70 if (pre_val < neg_threshold || val > pre_val) { 71 val = INT_MIN; 72 stage = 3; 73 continue; 74 } 75 } 76 pos++; // next character 77 break; 78 } 79 } 80 return val; 81 } 82 };
我是有多无聊写那么长，还是不能滥用状态的，写个简化的：
class Solution { public: int myAtoi(string str) { int res = 0; bool neg = false; int len = str.size(); int stage = 0; int pos = 0; char ch = 0; // skip white spaces while (pos < len && str[pos] == ' ') { pos++; } // try read sign if (pos < len && (str[pos] == '-' || str[pos] == '+')) { neg = str[pos++] == '-'; } // read digits while (pos < len && str[pos] >= '0' && str[pos] <= '9') { int old = res; if (neg) { res = res * 10 - (str[pos] - '0'); if (res > old || old < INT_MIN/10) { res = INT_MIN; break; } } else { res = res * 10 + str[pos] - '0'; if (res < old || old > INT_MAX/10) { res = INT_MAX; break; } } pos++; } return res; } };
