leetcode-master/problems/面试题02.07.链表相交.md

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# 面试题 02.07. 链表相交 

同：160.链表相交

[力扣题目链接](https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/)

给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 null 。

图示两个链表在节点 c1 开始相交：

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png)

题目数据 保证 整个链式结构中不存在环。

注意，函数返回结果后，链表必须 保持其原始结构 。

示例 1：

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png)

示例 2：

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png)

示例 3：

![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)


## 思路


简单来说，就是求两个链表交点节点的**指针**。 这里同学们要注意，交点不是数值相等，而是指针相等。

为了方便举例，假设节点元素数值相等，则节点指针相等。

看如下两个链表，目前curA指向链表A的头结点，curB指向链表B的头结点：

![面试题02.07.链表相交_1](https://code-thinking.cdn.bcebos.com/pics/面试题02.07.链表相交_1.png)

我们求出两个链表的长度，并求出两个链表长度的差值，然后让curA移动到，和curB 末尾对齐的位置，如图：

![面试题02.07.链表相交_2](https://code-thinking.cdn.bcebos.com/pics/面试题02.07.链表相交_2.png)

此时我们就可以比较curA和curB是否相同，如果不相同，同时向后移动curA和curB，如果遇到curA == curB，则找到交点。

否则循环退出返回空指针。

C++代码如下：

```CPP
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* curA = headA;
        ListNode* curB = headB;
        int lenA = 0, lenB = 0;
        while (curA != NULL) { // 求链表A的长度
            lenA++;
            curA = curA->next;
        }
        while (curB != NULL) { // 求链表B的长度
            lenB++;
            curB = curB->next;
        }
        curA = headA;
        curB = headB;
        // 让curA为最长链表的头，lenA为其长度
        if (lenB > lenA) {
            swap (lenA, lenB);
            swap (curA, curB);
        }
        // 求长度差
        int gap = lenA - lenB;
        // 让curA和curB在同一起点上（末尾位置对齐）
        while (gap--) {
            curA = curA->next;
        }
        // 遍历curA 和 curB，遇到相同则直接返回
        while (curA != NULL) {
            if (curA == curB) {
                return curA;
            }
            curA = curA->next;
            curB = curB->next;
        }
        return NULL;
    }
};
```

* 时间复杂度：O(n + m)
* 空间复杂度：O(1)

## 其他语言版本


### Java
```Java
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode curA = headA;
        ListNode curB = headB;
        int lenA = 0, lenB = 0;
        while (curA != null) { // 求链表A的长度
            lenA++;
            curA = curA.next;
        }
        while (curB != null) { // 求链表B的长度
            lenB++;
            curB = curB.next;
        }
        curA = headA;
        curB = headB;
        // 让curA为最长链表的头，lenA为其长度
        if (lenB > lenA) {
            //1. swap (lenA, lenB);
            int tmpLen = lenA;
            lenA = lenB;
            lenB = tmpLen;
            //2. swap (curA, curB);
            ListNode tmpNode = curA;
            curA = curB;
            curB = tmpNode;
        }
        // 求长度差
        int gap = lenA - lenB;
        // 让curA和curB在同一起点上（末尾位置对齐）
        while (gap-- > 0) {
            curA = curA.next;
        }
        // 遍历curA 和 curB，遇到相同则直接返回
        while (curA != null) {
            if (curA == curB) {
                return curA;
            }
            curA = curA.next;
            curB = curB.next;
        }
        return null;
    }

}
```

### Python
```python
（版本一）求长度，同时出发
class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        lenA, lenB = 0, 0
        cur = headA
        while cur:         # 求链表A的长度
            cur = cur.next 
            lenA += 1
        cur = headB 
        while cur:         # 求链表B的长度
            cur = cur.next 
            lenB += 1
        curA, curB = headA, headB
        if lenA > lenB:     # 让curB为最长链表的头，lenB为其长度
            curA, curB = curB, curA
            lenA, lenB = lenB, lenA 
        for _ in range(lenB - lenA):  # 让curA和curB在同一起点上（末尾位置对齐）
            curB = curB.next 
        while curA:         #  遍历curA 和 curB，遇到相同则直接返回
            if curA == curB:
                return curA
            else:
                curA = curA.next 
                curB = curB.next
        return None 
```
```python
（版本二）求长度，同时出发 （代码复用）
class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        lenA = self.getLength(headA)
        lenB = self.getLength(headB)
        
        # 通过移动较长的链表，使两链表长度相等
        if lenA > lenB:
            headA = self.moveForward(headA, lenA - lenB)
        else:
            headB = self.moveForward(headB, lenB - lenA)
        
        # 将两个头向前移动，直到它们相交
        while headA and headB:
            if headA == headB:
                return headA
            headA = headA.next
            headB = headB.next
        
        return None
    
    def getLength(self, head: ListNode) -> int:
        length = 0
        while head:
            length += 1
            head = head.next
        return length
    
    def moveForward(self, head: ListNode, steps: int) -> ListNode:
        while steps > 0:
            head = head.next
            steps -= 1
        return head
```
```python
（版本三）求长度，同时出发 （代码复用 + 精简）
class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        dis = self.getLength(headA) - self.getLength(headB)
        
        # 通过移动较长的链表，使两链表长度相等
        if dis > 0:
            headA = self.moveForward(headA, dis)
        else:
            headB = self.moveForward(headB, abs(dis))
        
        # 将两个头向前移动，直到它们相交
        while headA and headB:
            if headA == headB:
                return headA
            headA = headA.next
            headB = headB.next
        
        return None
    
    def getLength(self, head: ListNode) -> int:
        length = 0
        while head:
            length += 1
            head = head.next
        return length
    
    def moveForward(self, head: ListNode, steps: int) -> ListNode:
        while steps > 0:
            head = head.next
            steps -= 1
        return head
```
```python
（版本四）等比例法
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        # 处理边缘情况
        if not headA or not headB:
            return None
        
        # 在每个链表的头部初始化两个指针
        pointerA = headA
        pointerB = headB
        
        # 遍历两个链表直到指针相交
        while pointerA != pointerB:
            # 将指针向前移动一个节点
            pointerA = pointerA.next if pointerA else headB
            pointerB = pointerB.next if pointerB else headA
        
        # 如果相交，指针将位于交点节点，如果没有交点，值为None
        return pointerA
```
### Go

```go
func getIntersectionNode(headA, headB *ListNode) *ListNode {
    curA := headA
    curB := headB
    lenA, lenB := 0, 0
    // 求A，B的长度
    for curA != nil {
        curA = curA.Next
        lenA++
    }
    for curB != nil {
        curB = curB.Next
        lenB++
    }
    var step int
    var fast, slow *ListNode
    // 请求长度差，并且让更长的链表先走相差的长度
    if lenA > lenB {
        step = lenA - lenB
        fast, slow = headA, headB
    } else {
        step = lenB - lenA
        fast, slow = headB, headA
    }
    for i:=0; i < step; i++ {
        fast = fast.Next
    }
    // 遍历两个链表遇到相同则跳出遍历
    for fast != slow {
        fast = fast.Next
        slow = slow.Next
    }
    return fast
}
```

双指针

```go
func getIntersectionNode(headA, headB *ListNode) *ListNode {
    l1,l2 := headA, headB
    for l1 != l2 {
        if l1 != nil {
            l1 = l1.Next
        } else {
            l1 = headB
        }

        if l2 != nil {
            l2 = l2.Next
        } else {
            l2 = headA
        }
    }

    return l1
}
```

### javaScript

```js
var getListLen = function(head) {
    let len = 0, cur = head;
    while(cur) {
       len++;
       cur = cur.next;
    }
    return len;
}
var getIntersectionNode = function(headA, headB) {
    let curA = headA,curB = headB,
        lenA = getListLen(headA),   // 求链表A的长度
        lenB = getListLen(headB);  
    if(lenA < lenB) {       // 让curA为最长链表的头，lenA为其长度
    
        // 交换变量注意加 “分号” ，两个数组交换变量在同一个作用域下时
        // 如果不加分号，下面两条代码等同于一条代码: [curA, curB] = [lenB, lenA]
        
        [curA, curB] = [curB, curA];
        [lenA, lenB] = [lenB, lenA];
    }
    let i = lenA - lenB;   // 求长度差
    while(i-- > 0) {       // 让curA和curB在同一起点上（末尾位置对齐）
        curA = curA.next;
    }
    while(curA && curA !== curB) {  // 遍历curA 和 curB，遇到相同则直接返回
        curA = curA.next;
        curB = curB.next;
    }
    return curA;
};
```

TypeScript：

```typescript
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
  let sizeA: number = 0,
    sizeB: number = 0;
  let curA: ListNode | null = headA,
    curB: ListNode | null = headB;
  while (curA) {
    sizeA++;
    curA = curA.next;
  }
  while (curB) {
    sizeB++;
    curB = curB.next;
  }
  curA = headA;
  curB = headB;
  if (sizeA < sizeB) {
    [sizeA, sizeB] = [sizeB, sizeA];
    [curA, curB] = [curB, curA];
  }
  let gap = sizeA - sizeB;
  while (gap-- && curA) {
    curA = curA.next;
  }
  while (curA && curB) {
    if (curA === curB) {
      return curA;
    }
    curA = curA.next;
    curB = curB.next;
  }
  return null;
};
```

C：

```c
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    ListNode *l = NULL, *s = NULL;
    int lenA = 0, lenB = 0, gap = 0;
    // 求出两个链表的长度
    s = headA;
    while (s) {
        lenA ++;
        s = s->next;
    }
    s = headB;
    while (s) {
        lenB ++;
        s = s->next;
    }

    // 求出两个链表长度差
    if (lenA > lenB) {
        l = headA, s = headB;
        gap = lenA - lenB;
    } else {
        l = headB, s = headA;
        gap = lenB - lenA;
    }

    // 尾部对齐
    while (gap--) l = l->next;
    // 移动，并检查是否有相同的元素
    while (l) {
        if (l == s) return l;
        l = l->next, s = s->next;
    }

    return NULL;
}
```

Scala:
```scala
object Solution {
  def getIntersectionNode(headA: ListNode, headB: ListNode): ListNode = {
    var lenA = 0 // headA链表的长度
    var lenB = 0 // headB链表的长度
    var tmp = headA // 临时变量
    // 统计headA的长度
    while (tmp != null) {
      lenA += 1;
      tmp = tmp.next
    }
    // 统计headB的长度
    tmp = headB // 临时变量赋值给headB
    while (tmp != null) {
      lenB += 1
      tmp = tmp.next
    }
    // 因为传递过来的参数是不可变量，所以需要重新定义
    var listA = headA
    var listB = headB
    // 两个链表的长度差
    // 如果gap>0，lenA>lenB，headA(listA)链表往前移动gap步
    // 如果gap<0，lenA<lenB，headB(listB)链表往前移动-gap步
    var gap = lenA - lenB
    if (gap > 0) {
      // 因为不可以i-=1，所以可以使用for
      for (i <- 0 until gap) {
        listA = listA.next // 链表headA(listA) 移动
      }
    } else {
      gap = math.abs(gap) // 此刻gap为负值，取绝对值
      for (i <- 0 until gap) {
        listB = listB.next
      }
    }
    // 现在两个链表同时往前走，如果相等则返回
    while (listA != null && listB != null) {
      if (listA == listB) {
        return listA
      }
      listA = listA.next
      listB = listB.next
    }
    // 如果链表没有相交则返回null，return可以省略
    null
  }
}
```


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