feat: add python and java solutions to lcci question

添加《程序员面试金典》题解：面试题 03.01. 三合一

Author: yanglbme

lcci/03.01.Three in One/README.md

@@ -30,20 +30,85 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+二维数组解决；也可以使用一维数组，以下标 `0,3,6,..`、`1,4,7,..`、`2,5,8,..` 区分，一维数组最后三个元素记录每个栈的元素个数。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class TripleInOne:
+
+    def __init__(self, stackSize: int):
+        self._capacity = stackSize
+        self._s = [[], [], []]
+
+    def push(self, stackNum: int, value: int) -> None:
+        if len(self._s[stackNum]) < self._capacity:
+            self._s[stackNum].append(value)
+
+    def pop(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum].pop()
 
+    def peek(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum][-1]
+
+    def isEmpty(self, stackNum: int) -> bool:
+        return len(self._s[stackNum]) == 0
+
+
+# Your TripleInOne object will be instantiated and called as such:
+# obj = TripleInOne(stackSize)
+# obj.push(stackNum,value)
+# param_2 = obj.pop(stackNum)
+# param_3 = obj.peek(stackNum)
+# param_4 = obj.isEmpty(stackNum)
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class TripleInOne {
+    private int[] s;
+    private int capacity;
+
+    public TripleInOne(int stackSize) {
+        s = new int[stackSize * 3 + 3];
+        capacity = stackSize;
+    }
+    
+    public void push(int stackNum, int value) {
+        if (s[stackNum + 3 * capacity] < capacity) {
+            s[s[stackNum + 3 * capacity] * 3 + stackNum] = value;
+            ++s[stackNum + 3 * capacity];
+        }
+    }
+    
+    public int pop(int stackNum) {
+        if (isEmpty(stackNum)) {
+            return -1;
+        }
+        --s[stackNum + 3 * capacity];
+        return s[s[stackNum + 3 * capacity] * 3 + stackNum];
+    }
+    
+    public int peek(int stackNum) {
+        return isEmpty(stackNum) ? -1 : s[(s[stackNum + 3 * capacity] - 1) * 3 + stackNum];
+    }
+    
+    public boolean isEmpty(int stackNum) {
+        return s[stackNum + 3 * capacity] == 0;
+    }
+}
+
+/**
+ * Your TripleInOne object will be instantiated and called as such:
+ * TripleInOne obj = new TripleInOne(stackSize);
+ * obj.push(stackNum,value);
+ * int param_2 = obj.pop(stackNum);
+ * int param_3 = obj.peek(stackNum);
+ * boolean param_4 = obj.isEmpty(stackNum);
+ */
 ```
 
 ### ...

lcci/03.01.Three in One/README_EN.md

@@ -1,51 +1,142 @@
-# [03.01. Three in One](https://leetcode-cn.com/problems/three-in-one-lcci)
-
-## Description
-<p>Describe how you could use a single array to implement three stacks.</p>
-
-<p>Yout should implement&nbsp;<code>push(stackNum, value)</code>、<code>pop(stackNum)</code>、<code>isEmpty(stackNum)</code>、<code>peek(stackNum)</code>&nbsp;methods.&nbsp;<code>stackNum<font face="sans-serif, Arial, Verdana, Trebuchet MS">&nbsp;</font></code><font face="sans-serif, Arial, Verdana, Trebuchet MS">is the index of the stack.&nbsp;</font><code>value</code>&nbsp;is the value that pushed to the stack.</p>
-
-<p>The constructor requires a&nbsp;<code>stackSize</code>&nbsp;parameter, which represents the size of each stack.</p>
-
-<p><strong>Example1:</strong></p>
-
-<pre>
-<strong> Input</strong>: 
-[&quot;TripleInOne&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;isEmpty&quot;]
-[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
-<strong> Output</strong>: 
-[null, null, null, 1, -1, -1, true]
-<b>Explanation</b>: When the stack is empty, `pop, peek` return -1. When the stack is full, `push` does nothing.
-</pre>
-
-<p><strong>Example2:</strong></p>
-
-<pre>
-<strong> Input</strong>: 
-[&quot;TripleInOne&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;peek&quot;]
-[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
-<strong> Output</strong>: 
-[null, null, null, null, 2, 1, -1, -1]
-</pre>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [03.01. Three in One](https://leetcode-cn.com/problems/three-in-one-lcci)
+
+## Description
+<p>Describe how you could use a single array to implement three stacks.</p>
+
+
+
+<p>Yout should implement&nbsp;<code>push(stackNum, value)</code>、<code>pop(stackNum)</code>、<code>isEmpty(stackNum)</code>、<code>peek(stackNum)</code>&nbsp;methods.&nbsp;<code>stackNum<font face="sans-serif, Arial, Verdana, Trebuchet MS">&nbsp;</font></code><font face="sans-serif, Arial, Verdana, Trebuchet MS">is the index of the stack.&nbsp;</font><code>value</code>&nbsp;is the value that pushed to the stack.</p>
+
+
+
+<p>The constructor requires a&nbsp;<code>stackSize</code>&nbsp;parameter, which represents the size of each stack.</p>
+
+
+
+<p><strong>Example1:</strong></p>
+
+
+
+<pre>
+
+<strong> Input</strong>: 
+
+[&quot;TripleInOne&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;isEmpty&quot;]
+
+[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
+
+<strong> Output</strong>: 
+
+[null, null, null, 1, -1, -1, true]
+
+<b>Explanation</b>: When the stack is empty, `pop, peek` return -1. When the stack is full, `push` does nothing.
+
+</pre>
+
+
+
+<p><strong>Example2:</strong></p>
+
+
+
+<pre>
+
+<strong> Input</strong>: 
+
+[&quot;TripleInOne&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;peek&quot;]
+
+[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
+
+<strong> Output</strong>: 
+
+[null, null, null, null, 2, 1, -1, -1]
+
+</pre>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class TripleInOne:
+
+    def __init__(self, stackSize: int):
+        self._capacity = stackSize
+        self._s = [[], [], []]
+
+    def push(self, stackNum: int, value: int) -> None:
+        if len(self._s[stackNum]) < self._capacity:
+            self._s[stackNum].append(value)
+
+    def pop(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum].pop()
+
+    def peek(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum][-1]
+
+    def isEmpty(self, stackNum: int) -> bool:
+        return len(self._s[stackNum]) == 0
+
+
+# Your TripleInOne object will be instantiated and called as such:
+# obj = TripleInOne(stackSize)
+# obj.push(stackNum,value)
+# param_2 = obj.pop(stackNum)
+# param_3 = obj.peek(stackNum)
+# param_4 = obj.isEmpty(stackNum)
+```
+
+### Java
+
+```java
+class TripleInOne {
+    private int[] s;
+    private int capacity;
+
+    public TripleInOne(int stackSize) {
+        s = new int[stackSize * 3 + 3];
+        capacity = stackSize;
+    }
+    
+    public void push(int stackNum, int value) {
+        if (s[stackNum + 3 * capacity] < capacity) {
+            s[s[stackNum + 3 * capacity] * 3 + stackNum] = value;
+            ++s[stackNum + 3 * capacity];
+        }
+    }
+    
+    public int pop(int stackNum) {
+        if (isEmpty(stackNum)) {
+            return -1;
+        }
+        --s[stackNum + 3 * capacity];
+        return s[s[stackNum + 3 * capacity] * 3 + stackNum];
+    }
+    
+    public int peek(int stackNum) {
+        return isEmpty(stackNum) ? -1 : s[(s[stackNum + 3 * capacity] - 1) * 3 + stackNum];
+    }
+    
+    public boolean isEmpty(int stackNum) {
+        return s[stackNum + 3 * capacity] == 0;
+    }
+}
+
+/**
+ * Your TripleInOne object will be instantiated and called as such:
+ * TripleInOne obj = new TripleInOne(stackSize);
+ * obj.push(stackNum,value);
+ * int param_2 = obj.pop(stackNum);
+ * int param_3 = obj.peek(stackNum);
+ * boolean param_4 = obj.isEmpty(stackNum);
+ */
+```
+
+### ...
+```
+
+```

lcci/03.01.Three in One/Solution.java

@@ -0,0 +1,39 @@
+class TripleInOne {
+    private int[] s;
+    private int capacity;
+
+    public TripleInOne(int stackSize) {
+        s = new int[stackSize * 3 + 3];
+        capacity = stackSize;
+    }
+
+    public void push(int stackNum, int value) {
+        if (s[stackNum + 3 * capacity] < capacity) {
+            s[s[stackNum + 3 * capacity] * 3 + stackNum] = value;
+            ++s[stackNum + 3 * capacity];
+        }
+    }
+
+    public int pop(int stackNum) {
+        if (isEmpty(stackNum)) {
+            return -1;
+        }
+        --s[stackNum + 3 * capacity];
+        return s[s[stackNum + 3 * capacity] * 3 + stackNum];
+    }
+
+    public int peek(int stackNum) {
+        return isEmpty(stackNum) ? -1 : s[(s[stackNum + 3 * capacity] - 1) * 3 + stackNum];
+    }
+
+    public boolean isEmpty(int stackNum) {
+        return s[stackNum + 3 * capacity] == 0;
+    }
+}
+
+/**
+ * Your TripleInOne object will be instantiated and called as such: TripleInOne
+ * obj = new TripleInOne(stackSize); obj.push(stackNum,value); int param_2 =
+ * obj.pop(stackNum); int param_3 = obj.peek(stackNum); boolean param_4 =
+ * obj.isEmpty(stackNum);
+ */

lcci/03.01.Three in One/Solution.py

@@ -0,0 +1,26 @@
+class TripleInOne:
+
+    def __init__(self, stackSize: int):
+        self._capacity = stackSize
+        self._s = [[], [], []]
+
+    def push(self, stackNum: int, value: int) -> None:
+        if len(self._s[stackNum]) < self._capacity:
+            self._s[stackNum].append(value)
+
+    def pop(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum].pop()
+
+    def peek(self, stackNum: int) -> int:
+        return -1 if self.isEmpty(stackNum) else self._s[stackNum][-1]
+
+    def isEmpty(self, stackNum: int) -> bool:
+        return len(self._s[stackNum]) == 0
+
+
+# Your TripleInOne object will be instantiated and called as such:
+# obj = TripleInOne(stackSize)
+# obj.push(stackNum,value)
+# param_2 = obj.pop(stackNum)
+# param_3 = obj.peek(stackNum)
+# param_4 = obj.isEmpty(stackNum)