難度：4.5/5
一開始沒想出來好的解法而用中序轉後序爆開做，後來看到morris大神的解法才想出來可以把整個expression改成以及分隔的若干個expressions.
eval.c
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| #include <stdbool.h>
#include "eval.h"
bool check_invalid(int exp[]){
int n = exp[0];
for(int i = 2; i <= n; i += 2){
if(exp[i] > 4 || exp[i] <= 0) return true;
}
return false;
}
int eval(int exp[]) {
int n = exp[0];
if(check_invalid(exp)) return -2147483646;
int ans = 0;
int hold_val = exp[1];
for(int i = 3; i <= n; i += 2){
int op = exp[i - 1];
int num = exp[i];
if(op == 1){
ans += hold_val;
hold_val = num;
}
else if(op == 2){
ans += hold_val;
hold_val = -num;
}
else if(op == 3){
hold_val *= num;
}
else{
if(num == 0) return -2147483647;
else hold_val /= num;
}
}
ans += hold_val;
return ans;
}
C
|
eval.h