📝 add lc 86

DreamCats · DreamCats · commit bc1e67db9eab · 2021-04-10T13:27:50.000+08:00
diff --git a/Java/alg/README.md b/Java/alg/README.md
@@ -128,6 +128,7 @@
 - [78.子集](lc/78.子集.md)
 - [79.单词搜索](lc/79.单词搜索.md)
 - [82.删除排序链表中的重复元素2](lc/82.删除排序链表中的重复元素2.md)
+- [86.分隔链表](lc/86.分隔链表.md)
 
 ## 笔试
 
diff --git a/Java/alg/lc/86.分隔链表.md b/Java/alg/lc/86.分隔链表.md
@@ -0,0 +1,102 @@
+# 86. 分隔链表
+
+[url](https://leetcode-cn.com/problems/partition-list/))
+
+## 题目
+
+给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
+
+你应当 保留 两个分区中每个节点的初始相对位置。
+
+![](https://assets.leetcode.com/uploads/2021/01/04/partition.jpg)
+
+```
+输入：head = [1,4,3,2,5,2], x = 3
+输出：[1,2,2,4,3,5]
+输入：head = [2,1], x = 2
+输出：[1,2]
+```
+
+## 方法
+
+
+## code
+
+### js
+
+```js
+let partition = (head, x) => {
+    let dummy1, dummy2 = new ListNode(0);
+    let node1 = dummy1, node2 = dummy2;
+    while (head !== null) {
+        if (head.val < x) {
+            node1.next = head;
+            head = head.next;
+            node1 = node1.next;
+            node1.next = null;
+        } else {
+            node2.next = head;
+            head = head.next;
+            node2 = node2.next;
+            node2.next = null;
+        }
+    }
+    node1.nex = dummy2.next;
+    return dummy1.next;
+}
+```
+
+### go
+
+```go
+func partition(head *ListNode, x int) *ListNode {
+    if head == nil {return nil}
+    dummy1, dummy2 := &ListNode{}, &ListNode{}
+    cur1, cur2 := dummy1, dummy2
+    for head != nil {
+        if head.Val < x {
+            cur1.Next = head
+            head = head.Next
+            cur1 = cur1.Next
+            cur1.Next = nil
+        } else {
+            cur2.Next = head
+            head = head.Next
+            cur2 = cur2.Next
+            cur2.Next = nil
+        }
+    }
+    cur1.Next = dummy2.Next
+    return dummy1.Next
+}
+```
+
+
+
+### java
+
+```java
+class Solution {
+    public ListNode partition(ListNode head, int x) {
+        ListNode dummy1 = new ListNode(0);
+        ListNode dummy2 = new ListNode(0);
+        ListNode node1 = dummy1, node2 = dummy2;
+        while (head != null) {
+            if (head.val < x){
+                node1.next = head;
+                head = head.next;
+                node1 = node1.next;
+                node1.next = null;
+            } else {
+                node2.next = head;
+                head = head.next;
+                node2 = node2.next;
+                node2.next = null;
+            }
+        }
+        node1.next = dummy2.next;
+        return dummy1.next;
+    }
+}
+```
+
