Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
就是求逆波兰表达式（后续遍历）的结果。
1、直接求解，很慢
public class Solution {
public int evalRPN(String[] tokens) {
int len = tokens.length;
if( len == 0)
return 0;
for( int i = 0 ; i < len ; i ++ ){
if( tokens[i].equals("+") )
helper(tokens,i,1);
else if( tokens[i].equals("-") )
helper(tokens,i,2);
else if( tokens[i].equals("*") )
helper(tokens,i,3);
else if( tokens[i].equals("/") )
helper(tokens,i,4);
}
return Integer.valueOf(tokens[0]);
}
public void helper(String[] tokens,int pos,int type){
int pos1 = -1,pos2 = 0 ;
tokens[pos] = null;
while( pos >= 0 ){
if( tokens[pos] != null){
if( pos1 == -1)
pos1 = pos;
else{
pos2 = pos;
break;
}
}
pos--;
}
int num1 = Integer.valueOf(tokens[pos1]);
int num2 = Integer.valueOf(tokens[pos2]);
if( type == 1){
tokens[pos2] = String.valueOf(num1+num2);
}else if( type == 2){
tokens[pos2] = String.valueOf(num2-num1);
}else if( type == 3){
tokens[pos2] = String.valueOf(num2*num1);
}else if( type == 4){
tokens[pos2] = String.valueOf(num2/num1);
}
tokens[pos1] = null;
}
}
2、使用栈，很简单。
public class Solution {
public int evalRPN(String[] tokens) {
int len = tokens.length;
if( len == 0)
return 0;
Stack<Integer> stack = new Stack<Integer>();
for( int i = 0 ; i < len ; i ++ ){
if( tokens[i].equals("+") ){
int num1 = stack.pop();
int num2 = stack.pop();
stack.push(num1+num2);
}
else if( tokens[i].equals("-") ){
int num1 = stack.pop();
int num2 = stack.pop();
stack.push(num2-num1);
}
else if( tokens[i].equals("*") ){
int num1 = stack.pop();
int num2 = stack.pop();
stack.push(num1*num2);
}
else if( tokens[i].equals("/") ){
int num1 = stack.pop();
int num2 = stack.pop();
stack.push(num2/+num1);
}else{
stack.push(Integer.valueOf(tokens[i]));
}
}
return stack.pop();
}
}