Add solution 150

Author: yanglbme

README.md

@@ -58,6 +58,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 092 | [Reverse Linked List II](https://github.com/doocs/leetcode/tree/master/solution/092.Reverse%20Linked%20List%20II) | `Linked List` |
 | 094 | [Binary Tree Inorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/094.Binary%20Tree%20Inorder%20Traversal) | `Hash Table`, `Stack`, `Tree` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
+| 150 | [Evaluate Reverse Polish Notation](https://github.com/doocs/leetcode/tree/master/solution/150.Evaluate%20Reverse%20Polish%20Notation) | `Stack` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/doocs/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |
 
 

solution/150.Evaluate Reverse Polish Notation/README.md

@@ -0,0 +1,76 @@
+## 逆波兰表达式求值
+### 题目描述
+
+根据逆波兰表示法，求表达式的值。
+
+有效的运算符包括 `+`, `-`, `*`, `/` 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
+
+说明：
+
+整数除法只保留整数部分。
+给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
+示例 1：
+```
+输入: ["2", "1", "+", "3", "*"]
+输出: 9
+解释: ((2 + 1) * 3) = 9
+```
+
+示例 2：
+```
+输入: ["4", "13", "5", "/", "+"]
+输出: 6
+解释: (4 + (13 / 5)) = 6
+```
+
+示例 3：
+```
+输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
+输出: 22
+解释: 
+  ((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
+```
+
+### 解法
+遍历数组，若遇到操作数，将其压入栈中；若遇到操作符，从栈中弹出右操作数和左操作数，将运算结果压入栈中。最后栈中唯一的元素就是结果。
+
+
+```java
+class Solution {
+    public int evalRPN(String[] tokens) {
+        Stack<Integer> stack = new Stack<>();
+        for (String e : tokens) {
+            if (isNum(e)) {
+                stack.push(Integer.parseInt(e));
+            } else {
+                int y = stack.pop();
+                int x = stack.pop();
+                int z = 0;
+                switch (e) {
+                    case "+": z = x + y; break;
+                    case "-": z = x - y; break;
+                    case "*": z = x * y; break;
+                    case "/": z = x / y; break;    
+                }
+                stack.push(z);
+            }
+        }
+        return stack.peek();
+    }
+    
+    private boolean isNum(String val) {
+        try {
+            Integer.parseInt(val);
+            return true;
+        } catch (Exception e) {
+            return false;
+        }
+    }
+}
+```

solution/150.Evaluate Reverse Polish Notation/Solution.java

@@ -0,0 +1,31 @@
+class Solution {
+    public int evalRPN(String[] tokens) {
+        Stack<Integer> stack = new Stack<>();
+        for (String e : tokens) {
+            if (isNum(e)) {
+                stack.push(Integer.parseInt(e));
+            } else {
+                int y = stack.pop();
+                int x = stack.pop();
+                int z = 0;
+                switch (e) {
+                    case "+": z = x + y; break;
+                    case "-": z = x - y; break;
+                    case "*": z = x * y; break;
+                    case "/": z = x / y; break;    
+                }
+                stack.push(z);
+            }
+        }
+        return stack.peek();
+    }
+    
+    private boolean isNum(String val) {
+        try {
+            Integer.parseInt(val);
+            return true;
+        } catch (Exception e) {
+            return false;
+        }
+    }
+}