150. Evaluate Reverse Polish Notation

leetcode

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

Division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.
Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

解决思路

先初始化两个栈和一个队列，一个用来放数字，一个用来放符号。队列用来存放排好序的算式。
数字先进栈。碰到第一个符号，进符号的栈。然后按数字-符号-数字的顺序出栈并计算出表达式的值，将该值进队列。然后继续进栈，碰到第二个符号，如果队列不是空的，就将符号-数字进队列，计算出该表达式的值，往后以此类推。

第一个问题

输入的时候符号和数字混在一起，怎么实现输入？
好吧其实没有关系......一个char数组就搞定了
为了区分符号和数字（以下）

void judge(linklist* numnode,linklist *snode,char x){
    if(x=='+'||x=='-'||x=='*'||x=='/'){
        push(snode,x);
    }
    else{
        push(numnode,x);
    }
    return;
}
void ergodicAll(linklist* numnode,linklist *snode,char array[],int len){
    for(int i=0;i<len;i++){
        judge(numnode,snode,array[i]);
    }
    return;
}
void ergodic(linklist *numnode){
    while(!isEmpty(numnode)){
        printf("%d",numnode->data);
        numnode=numnode->next;
    }
    return;
}
void ergodic2(linklist *s){
    while(!isEmpty(s)){
        printf("%c",s->data);
        s=s->next;
    }
    return;
}