Add Solution for 142 Linked List Cycle II

cpp/142_Linked_List_Cycle_II.cpp

@@ -0,0 +1,76 @@
+// 142. Linked List Cycle II
+/**
+ * Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
+ * 
+ * Note: Do not modify the linked list.
+ *
+ * Follow up:
+ * Can you solve it without using extra space?
+ * 
+ * Tags: Linked List, Two Pointers
+ * 
+ * Similar Problems:  (E) Linked List Cycle, (M) Find the Duplicate Number
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+
+using namespace std;
+
+/**
+ * Definition for singly-linked list.
+ */
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+    ListNode(int x, ListNode *p) : val(x), next(p) {}
+};
+
+class Solution {
+public:
+    ListNode *detectCycle(ListNode *head) {
+        if (head == NULL || head->next == NULL) {
+            return NULL;
+        }
+
+        ListNode *slow = head, *fast = head;
+        while (fast != NULL && fast->next != NULL) {
+            slow = slow->next;
+            fast = fast->next->next;
+            // Suppose they first meet at step k, the length of the Cycle is r. 
+            // so 2k - k = nr, let n = 1, we have k = r
+            // m is the distance between start node of the cycle and first meeting node
+            // need (r - m) step to reach the start node of the cycle
+            // s is distance between the start node of list and the start node of cycle
+            // from the slow pointer: s = k - m
+            // ... - s - ... m ... - k
+            //       |               |
+            //      ...    r - m    ...
+            if (slow == fast) {
+                slow = head;
+                while (slow != fast) {
+                    slow = slow->next;
+                    fast = fast->next;
+                }
+
+                return slow;
+            }
+        }
+
+        return NULL;
+    }
+};
+
+int main() {
+    ListNode node5(5), node4(4, &node5), node3(3, &node4), node2(2, &node3), node1(1, &node2);
+    node5.next = &node3;
+    cout << &node3 << endl;
+    Solution sol;
+    ListNode *p = sol.detectCycle(&node1);
+    cout << p << endl;
+    cin.get();
+
+    return 0;
+}