feat: add solutions to lcof problem: No.25

Author: yanglbme

lcof/面试题25. 合并两个排序的链表/README.md

@@ -6,27 +6,22 @@
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
-
 class Solution {
 public:
     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
-        if (nullptr == l1 && nullptr == l2) {
-            return nullptr;
+        ListNode* dummy = new ListNode(0);
+        ListNode* cur = dummy;
+        while (l1 && l2) {
+            if (l1->val <= l2->val) {
+                cur->next = l1;
+                l1 = l1->next;
+            } else {
+                cur->next = l2;
+                l2 = l2->next;
+            }
+            cur = cur->next;
         }
-
-        if (nullptr == l1 || nullptr == l2) {
-            return l1 == nullptr ? l2 : l1;
-        }
-
-        ListNode* node = nullptr;
-        if (l1->val > l2->val) {
-            node = l2;
-            node->next = mergeTwoLists(l1, l2->next);
-        } else {
-            node = l1;
-            node->next = mergeTwoLists(l1->next, l2);
-        }
-
-        return node;
+        cur->next = l1 ? l1 : l2;
+        return dummy->next;
     }
 };

lcof/面试题25. 合并两个排序的链表/Solution.cpp

@@ -8,19 +8,19 @@
  */
 public class Solution {
     public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
-        ListNode p = new ListNode();
-        ListNode q = p;
+        ListNode dummy = new ListNode(0);
+        ListNode cur = dummy;
         while (l1 != null && l2 != null) {
-            if (l1.val < l2.val) {
-                p.next = l1;
+            if (l1.val <= l2.val) {
+                cur.next = l1;
                 l1 = l1.next;
             } else {
-                p.next = l2;
+                cur.next = l2;
                 l2 = l2.next;
             }
-            p = p.next;
+            cur = cur.next;
         }
-        p.next = l1 == null ? l2 : l1;
-        return q.next;
+        cur.next = l1 == null ? l2 : l1;
+        return dummy.next;
     }
 }

lcof/面试题25. 合并两个排序的链表/Solution.cs

@@ -1,14 +1,27 @@
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
 func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
-    if l1 == nil {
-        return l2
-    }
-    if l2 == nil {
-        return l1
-    }
-    if l1.Val <= l2.Val {
-        l1.Next = mergeTwoLists(l1.Next,l2)
-        return l1
-    }
-    l2.Next = mergeTwoLists(l1, l2.Next)
-    return l2
+	dummy := &ListNode{}
+	cur := dummy
+	for l1 != nil && l2 != nil {
+		if l1.Val <= l2.Val {
+			cur.Next = l1
+			l1 = l1.Next
+		} else {
+			cur.Next = l2
+			l2 = l2.Next
+		}
+		cur = cur.Next
+	}
+	if l1 == nil {
+		cur.Next = l2
+	} else {
+		cur.Next = l1
+	}
+	return dummy.Next
 }

lcof/面试题25. 合并两个排序的链表/Solution.go

@@ -9,18 +9,18 @@
 class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
         ListNode dummy = new ListNode(0);
-        ListNode p = dummy;
+        ListNode cur = dummy;
         while (l1 != null && l2 != null) {
             if (l1.val <= l2.val) {
-                p.next = l1;
+                cur.next = l1;
                 l1 = l1.next;
             } else {
-                p.next = l2;
+                cur.next = l2;
                 l2 = l2.next;
             }
-            p = p.next;
+            cur = cur.next;
         }
-        p.next = l1 == null ? l2 : l1;
+        cur.next = l1 == null ? l2 : l1;
         return dummy.next;
     }
 }

lcof/面试题25. 合并两个排序的链表/Solution.java

@@ -11,14 +11,18 @@
  * @return {ListNode}
  */
 var mergeTwoLists = function (l1, l2) {
-    if (!(l1 && l2)) {
-        return l1 || l2;
-    }
-    if (l1.val < l2.val) {
-        l1.next = mergeTwoLists(l1.next, l2);
-        return l1;
-    } else {
-        l2.next = mergeTwoLists(l2.next, l1);
-        return l2;
+    const dummy = new ListNode(0);
+    let cur = dummy;
+    while (l1 && l2) {
+        if (l1.val <= l2.val) {
+            cur.next = l1;
+            l1 = l1.next;
+        } else {
+            cur.next = l2;
+            l2 = l2.next;
+        }
+        cur = cur.next;
     }
+    cur.next = l1 || l2;
+    return dummy.next;
 };

lcof/面试题25. 合并两个排序的链表/Solution.js

@@ -7,15 +7,14 @@
 
 class Solution:
     def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
-        dummy = ListNode(0)
-        p = dummy
+        dummy = cur = ListNode(0)
         while l1 and l2:
             if l1.val <= l2.val:
-                p.next = l1
+                cur.next = l1
                 l1 = l1.next
             else:
-                p.next = l2
+                cur.next = l2
                 l2 = l2.next
-            p = p.next
-        p.next = l1 or l2
+            cur = cur.next
+        cur.next = l1 or l2
         return dummy.next

lcof/面试题25. 合并两个排序的链表/Solution.py

@@ -14,20 +14,18 @@ function mergeTwoLists(
     l1: ListNode | null,
     l2: ListNode | null,
 ): ListNode | null {
-    const duumy = new ListNode();
-    let cur = duumy;
+    const dummy = new ListNode(0);
+    let cur = dummy;
     while (l1 && l2) {
-        let node: ListNode;
-        if (l1.val < l2.val) {
-            node = l1;
+        if (l1.val <= l2.val) {
+            cur.next = l1;
             l1 = l1.next;
         } else {
-            node = l2;
+            cur.next = l2;
             l2 = l2.next;
         }
-        cur.next = node;
-        cur = node;
+        cur = cur.next;
     }
     cur.next = l1 || l2;
-    return duumy.next;
+    return dummy.next;
 }