Add solution 142 [Java]

Author: yanglbme

README.md

@@ -68,6 +68,7 @@ Complete [solutions](https://github.com/doocs/leetcode/tree/master/solution) to
 | 127 | [Word Ladder](https://github.com/doocs/leetcode/tree/master/solution/127.Word%20Ladder) | `Breadth-first Search` |
 | 130 | [Surrounded Regions](https://github.com/doocs/leetcode/tree/master/solution/130.Surrounded%20Regions) | `Depth-first Search`, `Breadth-first Search`, `Union Find` |
 | 137 | [Single Number II](https://github.com/doocs/leetcode/tree/master/solution/137.Single%20Number%20II) | `Bit Manipulation` |
+| 142 | [Linked List Cycle II](https://github.com/doocs/leetcode/tree/master/solution/142.Linked%20List%20Cycle%20II) | `Linked List`, `Two Pointers` |
 | 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/142.Linked List Cycle II/README.md

@@ -0,0 +1,71 @@
+## 环形链表 II
+### 题目描述
+
+给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 `null`。
+
+**说明：**不允许修改给定的链表。
+
+**进阶：**
+你是否可以不用额外空间解决此题？
+
+### 解法
+
+利用快慢指针，若快指针为 `null`，则不存在环，若快慢指针相遇，则存在环，此时退出循环，利用 `p1` 指向链表头结点，`p2` 指向快慢指针相遇点，`p1`, `p2` 同时前进，相遇时即为入环的第一个节点。
+
+**证明如下：**
+
+![solution142-slow-fast.png](http://p9ucdlghd.bkt.clouddn.com/solution142-slow-fast.png)
+
+假设链表到入环点距离为 `a`，入环点到快慢指针相遇点距离为 `b`，慢指针行程为`s`，快指针行程是它的 2 倍，相遇时快指针比慢指针多走了 `n * r`圈，即 `2s`。则：
+```
+①  a + b = s
+②  2s - s = n * r
+-> a + b = n * r
+-> a = n * r - b
+     = (n - 1) * r + r - b     
+```
+`r - b` 为相遇点到入环点的距离。
+
+`p1`, `p2` 同时向前走 `r - b`，`p2` 到达入环点，而 `p1` 距离入环点还有 `(n - 1) * r`，双方同时走 `(n - 1)` 圈即可相遇，此时相遇点就是入环点!
+
+
+```java
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode detectCycle(ListNode head) {
+        ListNode slow = head;
+        ListNode fast = head;
+        boolean hasCycle = false;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) {
+                hasCycle = true;
+                break;
+            }
+        }
+        
+        if (hasCycle) {
+            ListNode p1 = head;
+            ListNode p2 = slow;
+            while (p1 != p2) {
+                p1 = p1.next;
+                p2 = p2.next;
+            }
+            return p1;
+        }
+        return null;
+        
+    }
+}
+```

solution/142.Linked List Cycle II/Solution.java

@@ -0,0 +1,38 @@
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode detectCycle(ListNode head) {
+        ListNode slow = head;
+        ListNode fast = head;
+        boolean hasCycle = false;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) {
+                hasCycle = true;
+                break;
+            }
+        }
+        
+        if (hasCycle) {
+            ListNode p1 = head;
+            ListNode p2 = slow;
+            while (p1 != p2) {
+                p1 = p1.next;
+                p2 = p2.next;
+            }
+            return p1;
+        }
+        return null;
+        
+    }
+}