feat: add solutions to lcci question: 03.05.Sort of Stacks

添加《程序员面试金典》题解：面试题 03.05. 栈排序

Author: yanglbme

lcci/03.05.Sort of Stacks/README.md

@@ -31,20 +31,80 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+利用辅助栈实现 `push` 操作，其余操作不变。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class SortedStack:
+
+    def __init__(self):
+        self.s = []
+
+    def push(self, val: int) -> None:
+        t = []
+        while not self.isEmpty() and self.s[-1] < val:
+            t.append(self.s.pop())
+        self.s.append(val)
+        while len(t) > 0:
+            self.s.append(t.pop())
 
+    def pop(self) -> None:
+        if not self.isEmpty():
+            self.s.pop()
+
+    def peek(self) -> int:
+        return -1 if self.isEmpty() else self.s[-1]
+
+    def isEmpty(self) -> bool:
+        return len(self.s) == 0
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class SortedStack {
+    private Stack<Integer> s;
+    public SortedStack() {
+        s = new Stack<>();
+    }
+    
+    public void push(int val) {
+        Stack<Integer> t = new Stack<>();
+        while (!isEmpty() && s.peek() < val) {
+            t.push(s.pop());
+        }
+        s.push(val);
+        while (!t.isEmpty()) {
+            s.push(t.pop());
+        }
+    }
+    
+    public void pop() {
+        if (!isEmpty()) {
+            s.pop();
+        }
+    }
+    
+    public int peek() {
+        return isEmpty() ? -1 : s.peek();
+    }
+    
+    public boolean isEmpty() {
+        return s.isEmpty();
+    }
+}
+
+/**
+ * Your SortedStack object will be instantiated and called as such:
+ * SortedStack obj = new SortedStack();
+ * obj.push(val);
+ * obj.pop();
+ * int param_3 = obj.peek();
+ * boolean param_4 = obj.isEmpty();
+ */
 ```
 
 ### ...

lcci/03.05.Sort of Stacks/README_EN.md

@@ -1,52 +1,139 @@
-# [03.05. Sort of Stacks](https://leetcode-cn.com/problems/sort-of-stacks-lcci)
-
-## Description
-<p>Write a program to sort a stack such that the smallest items are on the top. You can use an additional temporary stack, but you may not copy the elements into any other data structure (such as an array). The stack supports the following operations: <code>push</code>, <code>pop</code>, <code>peek</code>, and <code>isEmpty</code>. When the stack is empty, <code>peek</code> should return -1.</p>
-
-<p><strong>Example1:</strong></p>
-
-<pre>
-<strong> Input</strong>: 
-[&quot;SortedStack&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;peek&quot;]
-[[], [1], [2], [], [], []]
-<strong> Output</strong>: 
-[null,null,null,1,null,2]
-</pre>
-
-<p><strong>Example2:</strong></p>
-
-<pre>
-<strong> Input</strong>:  
-[&quot;SortedStack&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;isEmpty&quot;]
-[[], [], [], [1], [], []]
-<strong> Output</strong>: 
-[null,null,null,null,null,true]
-</pre>
-
-<p><strong>Note:</strong></p>
-
-<ol>
-	<li>The total number of elements in the stack is within the range [0, 5000].</li>
-</ol>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [03.05. Sort of Stacks](https://leetcode-cn.com/problems/sort-of-stacks-lcci)
+
+## Description
+<p>Write a program to sort a stack such that the smallest items are on the top. You can use an additional temporary stack, but you may not copy the elements into any other data structure (such as an array). The stack supports the following operations: <code>push</code>, <code>pop</code>, <code>peek</code>, and <code>isEmpty</code>. When the stack is empty, <code>peek</code> should return -1.</p>
+
+
+
+<p><strong>Example1:</strong></p>
+
+
+
+<pre>
+
+<strong> Input</strong>: 
+
+[&quot;SortedStack&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;peek&quot;]
+
+[[], [1], [2], [], [], []]
+
+<strong> Output</strong>: 
+
+[null,null,null,1,null,2]
+
+</pre>
+
+
+
+<p><strong>Example2:</strong></p>
+
+
+
+<pre>
+
+<strong> Input</strong>:  
+
+[&quot;SortedStack&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;isEmpty&quot;]
+
+[[], [], [], [1], [], []]
+
+<strong> Output</strong>: 
+
+[null,null,null,null,null,true]
+
+</pre>
+
+
+
+<p><strong>Note:</strong></p>
+
+
+
+<ol>
+
+	<li>The total number of elements in the stack is within the range [0, 5000].</li>
+
+</ol>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class SortedStack:
+
+    def __init__(self):
+        self.s = []
+
+    def push(self, val: int) -> None:
+        t = []
+        while not self.isEmpty() and self.s[-1] < val:
+            t.append(self.s.pop())
+        self.s.append(val)
+        while len(t) > 0:
+            self.s.append(t.pop())
+
+    def pop(self) -> None:
+        if not self.isEmpty():
+            self.s.pop()
+
+    def peek(self) -> int:
+        return -1 if self.isEmpty() else self.s[-1]
+
+    def isEmpty(self) -> bool:
+        return len(self.s) == 0
+```
+
+### Java
+
+```java
+class SortedStack {
+    private Stack<Integer> s;
+    public SortedStack() {
+        s = new Stack<>();
+    }
+    
+    public void push(int val) {
+        Stack<Integer> t = new Stack<>();
+        while (!isEmpty() && s.peek() < val) {
+            t.push(s.pop());
+        }
+        s.push(val);
+        while (!t.isEmpty()) {
+            s.push(t.pop());
+        }
+    }
+    
+    public void pop() {
+        if (!isEmpty()) {
+            s.pop();
+        }
+    }
+    
+    public int peek() {
+        return isEmpty() ? -1 : s.peek();
+    }
+    
+    public boolean isEmpty() {
+        return s.isEmpty();
+    }
+}
+
+/**
+ * Your SortedStack object will be instantiated and called as such:
+ * SortedStack obj = new SortedStack();
+ * obj.push(val);
+ * obj.pop();
+ * int param_3 = obj.peek();
+ * boolean param_4 = obj.isEmpty();
+ */
+```
+
+### ...
+```
+
+```

lcci/03.05.Sort of Stacks/Solution.java

@@ -0,0 +1,40 @@
+class SortedStack {
+    private Stack<Integer> s;
+    public SortedStack() {
+        s = new Stack<>();
+    }
+    
+    public void push(int val) {
+        Stack<Integer> t = new Stack<>();
+        while (!isEmpty() && s.peek() < val) {
+            t.push(s.pop());
+        }
+        s.push(val);
+        while (!t.isEmpty()) {
+            s.push(t.pop());
+        }
+    }
+    
+    public void pop() {
+        if (!isEmpty()) {
+            s.pop();
+        }
+    }
+    
+    public int peek() {
+        return isEmpty() ? -1 : s.peek();
+    }
+    
+    public boolean isEmpty() {
+        return s.isEmpty();
+    }
+}
+
+/**
+ * Your SortedStack object will be instantiated and called as such:
+ * SortedStack obj = new SortedStack();
+ * obj.push(val);
+ * obj.pop();
+ * int param_3 = obj.peek();
+ * boolean param_4 = obj.isEmpty();
+ */

lcci/03.05.Sort of Stacks/Solution.py

@@ -0,0 +1,22 @@
+class SortedStack:
+
+    def __init__(self):
+        self.s = []
+
+    def push(self, val: int) -> None:
+        t = []
+        while not self.isEmpty() and self.s[-1] < val:
+            t.append(self.s.pop())
+        self.s.append(val)
+        while len(t) > 0:
+            self.s.append(t.pop())
+
+    def pop(self) -> None:
+        if not self.isEmpty():
+            self.s.pop()
+
+    def peek(self) -> int:
+        return -1 if self.isEmpty() else self.s[-1]
+
+    def isEmpty(self) -> bool:
+        return len(self.s) == 0