[LeetCode] Intersection of Two Linked Lists

★ 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;
}