合并两个有序的单链表，LeetCode-21

Author: fengxiaole

2-19.合并两个有序的单链表.cpp

@@ -0,0 +1,93 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+struct Node
+{
+    int val;
+    Node *next;
+    Node(int v) : val(v) { next = NULL; }
+};
+
+Node* Merge(Node *head1, Node *head2)
+{
+    if (head1 == NULL) return head2;
+    if (head2 == NULL) return head1;
+
+    Node *head = NULL;
+    Node *pre = NULL;
+    Node *p1 = head1;
+    Node *p2 = head2;
+    if (p1->val > p2->val)
+    {
+        head = p2;
+        pre = p2;
+        p2 = p2->next;
+    }
+    else
+    {
+        head = p1;
+        pre = p1;
+        p1 = p1->next;
+    }
+
+    while (NULL != p1 && NULL != p2)
+    {
+        if (p1->val > p2->val)
+        {
+            pre->next = p2;
+            pre = p2;
+            p2 = p2->next;
+        }
+        else
+        {
+            pre->next = p1;
+            pre = p1;
+            p1 = p1->next;
+        }
+    }
+
+    if (NULL != p1) pre->next = p1;
+    if (NULL != p2) pre->next = p2;
+
+    return head;
+}
+
+void PrintList(Node *head)
+{
+    Node *p = head;
+    while (NULL != p)
+    {
+        printf("%d ", p->val);
+        p = p->next;
+    }
+}
+int main() {
+    Node *node1_1 = new Node(1);
+    Node *node1_2 = new Node(2);
+    Node *node1_3 = new Node(3);
+    Node *node1_4 = new Node(4);
+    Node *node1_5 = new Node(5);
+
+    Node *node2_1 = new Node(0);
+    Node *node2_2 = new Node(3);
+    Node *node2_3 = new Node(4);
+    Node *node2_4 = new Node(7);
+    Node *node2_5 = new Node(10);
+
+    node1_1->next = node1_2;
+    node1_2->next = node1_3;
+    node1_3->next = node1_4;
+    node1_4->next = node1_5;
+
+    node2_1->next = node2_2;
+    node2_2->next = node2_3;
+    node2_3->next = node2_4;
+    node2_4->next = node2_5;
+
+    Node *head1 = node1_1;
+    Node *head2 = node2_1;
+    Node *head = Merge(head1, head2);
+    PrintList(head);
+    
+    return 0;
+}