Add Solution for 160 Intersection of Two Linked Lists

cpp/160_Intersection_of_Two_Linked_Lists.cpp

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+// 160. Intersection of Two Linked Lists
+/**
+ * Write a program to find the node at which the intersection of two singly linked lists begins.
+ * 
+ * For example, the following two linked lists:
+ * 
+ * A:          a1 - a2
+ *                    \
+ *                     c1 - c2 - c3
+ *                    /            
+ * B:     b1 - b2 - b3
+ * 
+ * begin to intersect at node c1.
+ * 
+ * Notes:
+ * 
+ *  If the two linked lists have no intersection at all, return null.
+ *  The linked lists must retain their original structure after the function returns.
+ *  You may assume there are no cycles anywhere in the entire linked structure.
+ *  Your code should preferably run in O(n) time and use only O(1) memory.
+ * 
+ * Tags: Linked List
+ * 
+ * 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) {}
+};
+
+// align the two list by point one's tail to the other's head
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        if (headA == NULL || headB == NULL) {
+            return NULL;
+        }
+
+        ListNode *pA = headA, *pB = headB;
+
+        while (pA != pB) {
+            // reset the pointer to the head of the other linked list at the end of first iteration
+            pA = pA ? pA->next : headB;
+            pB = pB ? pB->next : headA;
+        }
+
+        // make sure the two pointer traveled the same distance
+        // if the two lists have an intersection, the two pointer will meet at the intersection
+        // if not, they will meet at the end, i.e. NULL
+        return pA;
+    }
+};
+
+// align the start point by length difference
+class Solution_LenDifference {
+    int getLength(ListNode *head) {
+        int length = 0;
+        while (head != NULL) {
+            head = head->next;
+            length++;
+        }
+
+        return length;
+    }
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        if (headA == NULL || headB == NULL) {
+            return NULL;
+        }
+
+        int lenA = getLength(headA), lenB = getLength(headB);
+
+        // align the starting point
+        while (lenA > lenB) {
+            headA = headA->next;
+            lenA--;
+        }
+
+        while (lenB > lenA) {
+            headB = headB->next;
+            lenB--;
+        }
+
+        while (headA != headB) {
+            // compare the rest of the lists
+            headA = headA->next;
+            headB = headB->next;
+        }
+
+        // if there is no intersection, headA will go to null
+        return headA;
+    }
+};
+
+int main() {
+    ListNode c3(3), c2(2, &c3), c1(1, &c2);
+    ListNode a2(2, &c1), a1(1, &a2);
+    ListNode b3(3, &c1), b2(2, &b3), b1(1, &b2);
+    cout << &c1 << endl;
+
+    Solution sol;
+    ListNode *p = sol.getIntersectionNode(&a1, &b1);
+    cout << p << endl;
+
+    cin.get();
+
+    return 0;
+}