Leetcode刷题记——150. Evaluate Reverse Polish Notation(计算逆波兰表达式)
2016-12-26 22:34
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一、题目叙述:
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:
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二、解题思路:
这个问题就是求逆波兰式的值,非常简单,使用堆栈即可。
1、遇见数字,入栈。
2、遇见运算符,出栈得到两个操作数,运算后将结果入栈,最后栈中的数字即为结果。
但我不知道我刚开始为什么会超时。。。
三、源码:
import java.util.LinkedList;
public class Solution
{
public int evalRPN(String[] tokens)
{
LinkedList<Integer> stack = new LinkedList<Integer>();
int x, y, temp;
for (int i = 0; i < tokens.length; i ++)
{
if (tokens[i].equals("+") || tokens[i].equals("-") || tokens[i].equals("/") || tokens[i].equals("*"))
{
x = stack.removeFirst();
y = stack.removeFirst();
if (tokens[i].equals("+"))
stack.addFirst(y + x);
else if (tokens[i].equals("-"))
stack.addFirst(y - x);
else if (tokens[i].equals("*"))
stack.addFirst(y * x);
else if (tokens[i].equals("/"))
stack.addFirst(y / x);
}
else
{
temp = Integer.parseInt(tokens[i]);
stack.addFirst(temp);
}
}
// if (stack.size() == 1)
return stack.removeFirst();
}
public static void main(String[
4000
] args)
{
String[] a = {"0", "3", "/"};
Solution so = new Solution();
System.out.println(so.evalRPN(a));
}
}
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
Subscribe to see which companies asked this question
二、解题思路:
这个问题就是求逆波兰式的值,非常简单,使用堆栈即可。
1、遇见数字,入栈。
2、遇见运算符,出栈得到两个操作数,运算后将结果入栈,最后栈中的数字即为结果。
但我不知道我刚开始为什么会超时。。。
三、源码:
import java.util.LinkedList;
public class Solution
{
public int evalRPN(String[] tokens)
{
LinkedList<Integer> stack = new LinkedList<Integer>();
int x, y, temp;
for (int i = 0; i < tokens.length; i ++)
{
if (tokens[i].equals("+") || tokens[i].equals("-") || tokens[i].equals("/") || tokens[i].equals("*"))
{
x = stack.removeFirst();
y = stack.removeFirst();
if (tokens[i].equals("+"))
stack.addFirst(y + x);
else if (tokens[i].equals("-"))
stack.addFirst(y - x);
else if (tokens[i].equals("*"))
stack.addFirst(y * x);
else if (tokens[i].equals("/"))
stack.addFirst(y / x);
}
else
{
temp = Integer.parseInt(tokens[i]);
stack.addFirst(temp);
}
}
// if (stack.size() == 1)
return stack.removeFirst();
}
public static void main(String[
4000
] args)
{
String[] a = {"0", "3", "/"};
Solution so = new Solution();
System.out.println(so.evalRPN(a));
}
}
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