1669:合并两个链表

src/main/java/leetcode/editor/cn/MergeInBetweenLinkedLists.java

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+//给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。 
+//
+// 请你将 list1 中第 a 个节点到第 b 个节点删除，并将list2 接在被删除节点的位置。 
+//
+// 下图中蓝色边和节点展示了操作后的结果： 
+//
+// 请你返回结果链表的头指针。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//
+// 
+//输入：list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
+//输出：[0,1,2,1000000,1000001,1000002,5]
+//解释：我们删除 list1 中第三和第四个节点，并将 list2 接在该位置。上图中蓝色的边和节点为答案链表。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,100
+//0003,1000004]
+//输出：[0,1,1000000,1000001,1000002,1000003,1000004,6]
+//解释：上图中蓝色的边和节点为答案链表。
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 3 <= list1.length <= 104 
+// 1 <= a <= b < list1.length - 1 
+// 1 <= list2.length <= 104 
+// 
+// Related Topics 链表 
+// 👍 15 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.ListNode;
+
+//1669:合并两个链表
+public class MergeInBetweenLinkedLists {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new MergeInBetweenLinkedLists().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+
+    /**
+     * Definition for singly-linked list.
+     * public class ListNode {
+     * int val;
+     * ListNode next;
+     * ListNode() {}
+     * ListNode(int val) { this.val = val; }
+     * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+     * }
+     */
+    class Solution {
+        public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
+            ListNode temp = list1;
+            int num = 0;
+            while (temp != null) {
+                ListNode trans = new ListNode(0);
+                if (num == a - 1) {
+                    trans.next = temp.next;
+                    temp.next = list2;
+                    temp = trans;
+                }
+                if (num == b + 1) {
+                    ListNode temp2 = list2;
+                    while (temp2 != null && temp2.next != null) {
+                        temp2 = temp2.next;
+                    }
+                    assert temp2 != null;
+                    temp2.next = temp;
+                    break;
+                }
+                num++;
+                temp = temp.next;
+            }
+            return list1;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/MergeInBetweenLinkedLists.md

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+<p>给你两个链表 <code>list1</code> 和 <code>list2</code> ，它们包含的元素分别为 <code>n</code> 个和 <code>m</code> 个。</p>
+
+<p>请你将 <code>list1</code> 中第 <code>a</code> 个节点到第 <code>b</code> 个节点删除，并将<code>list2</code> 接在被删除节点的位置。</p>
+
+<p>下图中蓝色边和节点展示了操作后的结果：</p>
+<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/28/fig1.png" style="height: 130px; width: 504px;" />
+<p>请你返回结果链表的头指针。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/28/merge_linked_list_ex1.png" style="width: 406px; height: 140px;" /></p>
+
+<pre>
+<b>输入：</b>list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
+<b>输出：</b>[0,1,2,1000000,1000001,1000002,5]
+<b>解释：</b>我们删除 list1 中第三和第四个节点，并将 list2 接在该位置。上图中蓝色的边和节点为答案链表。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/28/merge_linked_list_ex2.png" style="width: 463px; height: 140px;" />
+<pre>
+<b>输入：</b>list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
+<b>输出：</b>[0,1,1000000,1000001,1000002,1000003,1000004,6]
+<b>解释：</b>上图中蓝色的边和节点为答案链表。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 <= list1.length <= 10<sup>4</sup></code></li>
+	<li><code>1 <= a <= b < list1.length - 1</code></li>
+	<li><code>1 <= list2.length <= 10<sup>4</sup></code></li>
+</ul>
+<div><div>Related Topics</div><div><li>链表</li></div></div>\n<div><li>👍 15</li><li>👎 0</li></div>