代码天涯路

160. 相交链表(Intersection of Two Linked Lists)E

英文题目

  • Write a program to find the node at which the intersection of two singly linked lists begins.

  • For example, the following two linked lists:

  • begin to intersect at node c1.

  • Example 1:

  • Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

  • Example 2:

  • Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

  • Example 3:

  • Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.

  • Explanation: The two lists do not intersect, so return null.

  • Notes:

  • If the two linked lists have no intersection at all, return null.

  • The linked lists must retain their original structure after the function returns.

  • You may assume there are no cycles anywhere in the entire linked structure.

  • Each value on each linked list is in the range [1, 10^9].

  • Your code should preferably run in O(n) time and use only O(1) memory.

中文题目

  • 编写一个程序，找到两个单链表相交的起始节点。

  • 如下面的两个链表**：**

  • 在节点 c1 开始相交。

  • 示例 1：

  • 输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
输出：Reference of the node with value = 8
输入解释：相交节点的值为 8 （注意，如果两个链表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [4,1,8,4,5]，链表 B 为 [5,0,1,8,4,5]。在 A 中，相交节点前有 2 个节点；在 B 中，相交节点前有 3 个节点。

  • 示例 2：

  • 输入：intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
输出：Reference of the node with value = 2
输入解释：相交节点的值为 2 （注意，如果两个链表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [0,9,1,2,4]，链表 B 为 [3,2,4]。在 A 中，相交节点前有 3 个节点；在 B 中，相交节点前有 1 个节点。

  • 示例 3：

  • 输入：intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
输出：null
输入解释：从各自的表头开始算起，链表 A 为 [2,6,4]，链表 B 为 [1,5]。由于这两个链表不相交，所以 intersectVal 必须为 0，而 skipA 和 skipB 可以是任意值。
解释：这两个链表不相交，因此返回 null。

  • 注意：

  • 如果两个链表没有交点，返回 null.

  • 在返回结果后，两个链表仍须保持原有的结构。

  • 可假定整个链表结构中没有循环。

  • 程序尽量满足 O(n) 时间复杂度，且仅用 O(1) 内存。

哈希集合

  • 用集合存储A链表所有节点，然后遍历B链表查看是否存在

  • 时间复杂度O(m+n)，空间复杂度O(m)，m,n 分别是链表headA和headB的长度

    # python3: 时间 132 ms, 击败 90.42%; 内存 31.2 MB, 击败 15.61%
# 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) -> Optional[ListNode]:
        visited = set()
        while headA:
            visited.add(headA)
            headA = headA.next
        while headB:
            if headB in visited:
                return headB
            headB = headB.next
        return None

    // c++: 时间 64 ms, 击败 8.48%; 内存 16.8 MB, 击败 12.89%
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        unordered_set<ListNode *> visited;
        while (headA != nullptr) {
            visited.insert(headA);
            headA = headA->next;
        }
        while (headB != nullptr) {
            if (visited.find(headB) != visited.end()) {
                return headB;
            }
            headB = headB->next;
        }
        return nullptr;
    }
};

    // java: 时间 5 ms, 击败 20.38%; 内存 45.3 MB, 击败 48%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        Set<ListNode> visited = new HashSet<>();
        while (headA != null) {
            visited.add(headA);
            headA = headA.next;
        }
        while (headB != null) {
            if (visited.contains(headB)) {
                return headB;
            }
            headB = headB.next;
        }
        return null;
    }
}

    // go: 时间 32 ms, 击败 34.30%; 内存 7.3 MB, 击败 5.3%
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
    visited := map[*ListNode]bool{}
    for headA != nil {
        visited[headA] = true
        headA = headA.Next
    }
    for headB != nil {
        if _, ok := visited[headB]; ok {
            return headB;
        }
        headB = headB.Next
    }
    return nil
}

    // javascript: 时间 80 ms, 击败 87.90%; 内存 49 MB, 击败 24%
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
    const visited = new Set();
    while (headA != null) {
        visited.add(headA);
        headA = headA.next;
    }
    while (headB != null) {
        if (visited.has(headB)) {
            return headB
        }
        headB = headB.next;
    }
    return null;
};

双指针

  • 思路是：先遍历完自己的链表，再遍历对方的链表

  • 如图，设「第一个公共节点」是node

  • 指针 pA 先遍历完链表 headA ，再开始遍历链表 headB ，当走到 node 时，共走步数为：a + (b - c)

  • 指针 pB 先遍历完链表 headB ，再开始遍历链表 headA ，当走到 node 时，共走步数为：b + (a - c)

  • 此时指针 pA，pB 重合，有两种情况：

  • 1.若两链表有公共尾部(即 c > 0)：指针 pA，pB 同时指向「第一个公共节点」node

  • 2.若两链表无公共尾部(即 c = 0)：指针 pA，pB 同时指向 null

  • 此时返回 pA 即可(pB也行)，它们相等

  • 时间复杂度O(a+b)，空间复杂度O(1)

    # python3: 时间 144 ms, 击败 54.79%; 内存 31 MB, 击败 25.98%
# 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:
        if not headA or not headB:
            return None
        pA, pB = headA, headB
        # 如果pA,pB当前的节点是存在的，往下走
        # 这里还用了个特性，None != None 时，结果是False
        while pA != pB:
            pA = pA.next if pA else headB
            pB = pB.next if pB else headA
        return pA

    // c++: 时间 36 ms, 击败 90.22%; 内存 14.3 MB, 击败 40.67%
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == nullptr || headB == nullptr) {
            return nullptr;
        }
        ListNode *pA = headA, *pB = headB;
        while (pA != pB) {
            pA = pA != nullptr ? pA->next : headB;
            pB = pB != nullptr ? pB->next : headA;
        }
        return pA;
    }
};

    // java: 时间 1 ms, 击败 98.37%; 内存 45.2 MB, 击败 63.38%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        ListNode pA = headA, pB = headB;
        while (pA != pB) {
            pA = pA != null ? pA.next : headB;
            pB = pB != null ? pB.next : headA;
        }
        return pA;
    }
}

    // go: 时间 28 ms, 击败 74.36%; 内存 6.5 MB, 击败 65.75%
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
    if headA == nil || headB == nil {
        return nil
    }
    pA, pB := headA, headB
    for pA != pB {
        if pA != nil {
            pA = pA.Next
        } else {
            pA = headB
        }
        if pB != nil {
            pB = pB.Next
        } else {
            pB = headA
        }
    }
    return pA
}

    // javascript: 时间 84 ms, 击败 76.76%; 内存 48.1 MB, 击败 89.31%
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
    if (headA == null || headB == null) {
        return null;
    }
    let pA = headA, pB = headB;
    while (pA != pB) {
        pA = pA != null ? pA.next : headB;
        pB = pB != null ? pB.next : headA;
    }
    return pA;
};
Last Updated:
Contributors: Shiqi Lu