[150] Evaluate Reverse Polish Nottion

這題目.....根本就是某考題的原型啊  XD！

看來leetcode上的比較簡單，因為它有限制比較多的input

總之

其實比完是不是加減乘除之後可以直接做運算了

就不用在程式的下段又來一整組的if else. 

就降子 XD

int evalRPN(char ** tokens, int tokensSize){

int head = -1;

int *stack = calloc (10000, sizeof(int));

for (int i=0;i<tokensSize; i++)

{

int ops = 0; // +,-,*,/ = 1,2,3,4

if (strcmp(tokens[i],"+")==0)

ops=1;

else if (strcmp(tokens[i],"-")==0)

ops=2;

else if (strcmp(tokens[i],"*")==0)

ops=3;

else if (strcmp(tokens[i],"/")==0)

ops=4;

if (ops==0)

{

stack[++head] = atoi(tokens[i]);

}

else

{

if (head<1)

return -1;

int val1, val2, result;

val1= stack[head-1];

val2= stack[head];

if (ops==1)

result = val1 + val2;

else if (ops==2)

result = val1 - val2;

else if (ops==3)

result = val1 * val2;

else if (ops==4)

result = val1 / val2;

stack[head-1]= result;

head--;

}

}

if (head>0)

return -1;

return stack[head];

}