说明
算法：Evaluate Reverse Polish Notation
LeetCode地址：https://leetcode.com/problems/evaluate-reverse-polish-notation/
题目：
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
Example 2:
Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:
Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22
解题思路
逆波兰表达式求值实际上是遇到加减乘除符号，跟符号最近的数字先计算。
如果是加号、乘号那么左右顺序无关，减号、除号则最近的数字在右边。
所以用栈就很好处理上面计算，数字入栈，遇到符号则把数字出栈。
时间复杂度为 O(N)。
代码实现
import java.util.Stack;
public class EvaluateReversePolishNotation {
public int evalRPN(String[] tokens) {
Stack<Integer> stack = new Stack<>();
int temp = 0;
for (String s: tokens) {
if (s.equals("+")) {
temp = stack.pop() + stack.pop();
} else if (s.equals("-")) {
int right = stack.pop();
int left = stack.pop();
temp = left - right;
} else if (s.equals("*")) {
temp = stack.pop() * stack.pop();
} else if (s.equals("/")) {
int right = stack.pop();
int left = stack.pop();
temp = left / right;
} else {
temp = Integer.parseInt(s);
}
stack.push(temp);
}
return temp;
}
public static void main(String[] args) {
EvaluateReversePolishNotation obj = new EvaluateReversePolishNotation();
String[] input = {"2", "1", "+", "3", "*"};
System.out.println("ouput: " + obj.evalRPN(input));
}
}
运行结果
ouput: 9
代码执行效率
Runtime: 6 ms, faster than 90.38% of Java online submissions for Evaluate Reverse Polish Notation.
Memory Usage: 37.1 MB, less than 33.65% of Java online submissions for Evaluate Reverse Polish Notation.
总结
考查栈的运用，小心减法、除法的数字位置问题。