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Author: assassint2017

Merge Two Sorted Lists/Merge_Two_Sorted_Lists.cpp

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+// 第一种思路，完全新建一个链表，不去破坏原始的两个小链表
+// 12ms 43.69%
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution 
+{
+public:
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
+    {   
+        ListNode *head = new ListNode(0), *rear = head;
+        
+        // 循环判断直到两个链表有一个为空
+        while (l1 && l2)
+        {
+            int minValue;
+            
+            if (l1->val <= l2->val)
+            {
+                minValue = l1->val;
+                l1 = l1->next;
+            }
+            else
+            {
+                minValue = l2->val;
+                l2 = l2->next;
+            }
+                
+            rear->next = new ListNode(minValue);
+            rear = rear->next;
+        }
+        
+        // 将剩下的链表接到总链表的后边
+        for (ListNode *rest = l1 ? l1 : l2;rest;rest = rest->next)
+        {
+            rear->next = new ListNode(rest->val);
+            rear = rear->next;
+        }
+        
+        return head->next;
+    }
+};
+
+// 第二种思路，破坏原始的两个小链表，将一个链表插入到另外一个链表中
+// 12ms 43.69%
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        
+        if (!l1) return l2;
+        if (!l2) return l1;
+        
+        ListNode *head;
+        if (l1->val > l2->val)
+        {
+            ListNode *temp = l1;
+            l1 = l2;
+            l2 = temp;                      
+        }
+        head = l1;
+        
+        // 循环插入直到两个链表有一个为空
+        while(l1 && l2)
+        {
+            while (l1->next && l1->next->val <= l2->val)
+                l1 = l1->next;
+            
+            if (!l1->next)break;
+            
+            ListNode *temp = l2;
+            l2 = l2->next;
+            
+            temp->next = l1->next;
+            l1->next = temp;
+            
+            l1 = l1->next;      
+        }
+        
+        // 将剩下的链表接到总链表的后边
+        if (!l1->next)
+            l1->next = l2; 
+        return head;
+    }
+};