algorithm-base/animation-simulation/链表篇/leetcode86分隔链表.md

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#### [86. 分隔链表](https://leetcode-cn.com/problems/partition-list/)

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://img-blog.csdnimg.cn/20210319190335143.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzMzODg1OTI0,size_16,color_FFFFFF,t_70)

示例 1：

输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]
示例 2：

输入：head = [2,1], x = 2
输出：[1,2]

来源：力扣（LeetCode）

这个题目我的做题思路是这样的，我们先创建一个侦察兵，侦察兵负责比较链表值和 x 值，如果 >= 的话则接在 big 链表上，小于则接到 small 链表上，最后一个细节就是我们的 big 链表尾部要加上 null，不然会形成环。这是这个题目的一个小细节，很重要。

中心思想就是，将链表先分后合。

下面我们来看模拟视频吧。希望能给各位带来一丢丢帮助。

![](https://img-blog.csdnimg.cn/20210319190417499.gif)

**题目代码**

Java Code:

```java
class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode pro = head;
        ListNode big = new ListNode(-1);
        ListNode small = new ListNode(-1);
        ListNode headbig = big;
        ListNode headsmall = small;
        //分
        while (pro != null) {
            //大于时，放到 big 链表上
            if (pro.val >= x) {
                big.next = pro;
                big = big.next;
            //小于时，放到 small 链表上
            }else {
                small.next = pro;
                small = small.next;
            }
            pro = pro.next;
        }
        //细节
        big.next = null;
        //合
        small.next = headbig.next;
        return headsmall.next;
    }
}
```

C++ Code:

```cpp
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode * pro = head;
        ListNode * big = new ListNode(-1);
        ListNode * small = new ListNode(-1);
        ListNode * headbig = big;
        ListNode * headsmall = small;
        //分
        while (pro != nullptr) {
            //大于时，放到 big 链表上
            if (pro->val >= x) {
                big->next = pro;
                big = big->next;
            //小于时，放到 small 链表上
            }else {
                small->next = pro;
                small = small->next;
            }
            pro = pro->next;
        }
        //细节
        big->next = nullptr;
        //合
        small->next = headbig->next;
        return headsmall->next;
    }
};
```

JS Code:

```js
var partition = function (head, x) {
  let pro = head;
  let big = new ListNode(-1);
  let small = new ListNode(-1);
  let headbig = big;
  let headsmall = small;
  //分
  while (pro) {
    //大于时，放到 big 链表上
    if (pro.val >= x) {
      big.next = pro;
      big = big.next;
      //小于时，放到 small 链表上
    } else {
      small.next = pro;
      small = small.next;
    }
    pro = pro.next;
  }
  //细节
  big.next = null;
  //合
  small.next = headbig.next;
  return headsmall.next;
};
```

Python Code:

```python
class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        pro = head
        big = ListNode(-1)
        small = ListNode(-1)
        headbig = big
        headsmall = small
        # 分
        while pro is not None:
            # 大于时，放到 big 链表上
            if pro.val >= x:
                big.next = pro
                big = big.next
            # 小于时，放到 small 链表上
            else:
                small.next = pro
                small = small.next
            pro = pro.next
        # 细节
        big.next = None
        # 合
        small.next = headbig.next
        return headsmall.next
```

Swift Code：

```swift
class Solution {
    func partition(_ head: ListNode?, _ x: Int) -> ListNode? {
        var pro = head
        var big = ListNode(-1)
        var small = ListNode(-1)
        var headbig = big
        var headsmall = small
        //分
        while pro != nil {
            //大于时，放到 big 链表上
            if pro!.val >= x {
                big.next = pro
                big = big.next!
            //小于时，放到 small 链表上
            } else {
                small.next = pro
                small = small.next!
            }
            pro = pro?.next
        }
        //细节
        big.next = nil
        //合
        small.next = headbig.next
        return headsmall.next
    }
}
```

Go Code:

```go
func partition(head *ListNode, x int) *ListNode {
    big, small := &ListNode{}, &ListNode{}
    headBig, headSmall := big, small
    temp := head
    for temp != nil {
        // 分开存
        if temp.Val < x {
            small.Next = temp
            small = small.Next
        } else {
            big.Next = temp
            big = big.Next
        }
        temp = temp.Next
    }
    // 最后一个节点指向nil
    big.Next = nil
    // 存小数的链表和存大数的连起来
    small.Next = headBig.Next
    return headSmall.Next
}
```