Add merge two sorted linked list function

Author: chachaxw

leetcode/linked_list/merge_two_sorted_list

@@ -0,0 +1,90 @@
+/**
+ * @Author: Chacha 
+ * @Date: 2019-01-06 22:50:09 
+ * @Last Modified by: Chacha
+ * @Last Modified time: 2019-01-06 23:13:13
+ */
+
+#include<iostream>
+#include<string>
+using namespace std;
+
+/** 
+ * Definition for singly-linked list(单向链表)
+ * Source: https://zh.wikipedia.org/wiki/链表
+ */
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+};
+
+class Solution {
+public:
+    /**
+     * Merge two sorted linked lists and return it as a new list.
+     * The new list should be made by splicing together the nodes of the first two lists.
+     * 
+     * Example: 
+     *  Given 1->3->8->11->15->null, 2->null
+     *  return 1->2->3->8->11->15->null
+     * 
+     * Source: https://leetcode.com/problems/merge-two-sorted-lists/
+     * 
+     * Operation steps:
+     *  1. Exception handling, included in dummy->next
+     *  2. Create the dummy and lastNode nodes, let the lastNode pointed to dummy node
+     *  3. Loop processing for non-empty l1, l2, linking the smaller of l1/l2 to lastNode->next, recursing lastNode
+     *  4. one of the linked lists in l1/l2 is empty to exit the while loop, 
+     *     and the non-empty list header is linked to lastNode->next
+     *  5. Return dummy->next, which is the final head pointer
+     */ 
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        ListNode* dummy = new ListNode(0);
+        ListNode* lastNode = dummy;
+        
+        while(l1 != NULL && l2 != NULL){
+
+            if (l1->val < l2->val) {
+                lastNode->next = l1;
+                l1 = l1->next;
+            } else {
+                lastNode->next = l2;
+                l2 = l2->next;
+            }
+
+            lastNode = lastNode->next;
+        }
+
+        lastNode->next = (l1 != NULL) ? l1 : l2;
+        return dummy->next;
+    }
+};
+
+/* Function to print nodes in a given linked list */
+void printList(ListNode *node) { 
+    while (node != NULL) { 
+        printf("%d ", node->val); 
+        node = node->next; 
+    }
+}
+
+int main() {
+    ListNode* l1 = new ListNode(1);
+    l1->next = new ListNode(3);
+    l1->next->next = new ListNode(5);
+    l1->next->next->next = new ListNode(7);
+    l1->next->next->next->next = new ListNode(9);
+
+    ListNode* l2 = new ListNode(2);
+    l2->next = new ListNode(4);
+    l2->next->next = new ListNode(6);
+    l2->next->next->next = new ListNode(8);
+    l2->next->next->next->next = new ListNode(10);
+
+    ListNode* merged = Solution().mergeTwoLists(l1, l2);
+    
+    printList(merged);
+
+    return 0;
+}