50017. Expression
難度：4.8/5
難但是不到做不出來。
要注意edge cases：operator減號跟負號的地方該怎麼決定。
Second Try: 3.5/5 Used Time: 20:25 Used Time: 11:011
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85#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include "expression.h"
#define err_num -2147483648
int find_op(char *string, int start, int finish, int* pos){
int step = 0;
int seen_int = 0;
for(int i = start + 1; i < finish - 1; i++){
char ch = string[i];
if(ch == '(') step++;
if(ch == ')') step--;
if(isdigit(ch)) seen_int = 1;
if(seen_int == 0) continue;
if(step < 0) return 0;
if(step == 0){
if(ch == '+'){
*pos = i;
return 1;
}
else if(ch == '-'){
*pos = i;
return 2;
}
else if(ch == '*'){
*pos = i;
return 3;
}
else if(ch == '/'){
*pos = i;
return 4;
}
}
}
return 0;
}
int expressionEval(char *string, int start, int finish){
if(isdigit(string[start])){ // case 1
if(start + 1 != finish) return err_num;
else return string[start] - '0';
}
else if(string[start] == '-'){ // case 2
return -(expressionEval(string, start + 1, finish));
}
else if(string[start] == '('){ // case 3, 4
int pos = -1;
int op_case = find_op(string, start, finish, &pos);
if(op_case != 0){
int f = expressionEval(string, start + 1, pos);
int s = expressionEval(string, pos + 1, finish - 1);
if(f == err_num || s == err_num) return err_num;
if(op_case == 1){ // +
return f + s;
}
else if(op_case == 2){ // -
return f - s;
}
else if(op_case == 3){ // *
return f * s;
}
else if(op_case == 4){ // /
if(s == 0) return err_num;
else return f / s;
}
}
else{
return err_num;
}
}
else{
return err_num;
}
}
int expression(char *string){
return expressionEval(string, 0, strlen(string));
}
Edge Case Input
1 | |
Edge Case Output
``` 8 -2 8 2 -8 8 -2