[LeetCode] Intersection of Two Linked Lists

寻找链表交点

最新推荐文章于 2025-03-18 05:59:11 发布

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最新推荐文章于 2025-03-18 05:59:11 发布

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本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过计算两个链表的长度并调整较长链表的起始位置，使得两链表剩余部分长度相等，然后同步遍历比较节点值来确定是否存在交点。

★ 题目

https://leetcode.com/problems/intersection-of-two-linked-lists/

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

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

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

★ 代码

42 / 42 test cases passed.
Runtime: 20 ms

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
#define rint register int
typedef struct ListNode ListNode;

int getSize(ListNode* head) {
    rint len = 0;
    ListNode* curr = head;
    while (curr != NULL) {
        len++;
        curr = curr->next;
    }
    return len;
}

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if (headA == NULL || headB == NULL) {
        return NULL;
    }

    int lenA = getSize(headA);
    int lenB = getSize(headB);
    int diff = (lenA > lenB)?(lenA - lenB):(lenB - lenA);

    ListNode* curr = NULL;// 指向较短的链表
    ListNode* currLong = NULL;// 指向较长的链表
    if (lenA > lenB) {
        currLong = headA;
        curr = headB;
    } else {
        currLong = headB;
        curr = headA;
    }
    rint i = 0;
    while (i < diff) {
        currLong = currLong->next;
        i++;
    }

    bool found = false;
    while (curr != NULL && currLong != NULL) {
        if (currLong->val == curr->val) {
            found = true;
            break;
        }
        currLong = currLong->next;
        curr = curr->next;
    }
    return found?curr:NULL;
}