LEETCODE-Intersection of Two Linked Lists

本文介绍了一种寻找两个单链表开始相交节点的方法。针对不同长度的链表,通过计算差值并调整遍历起点,确保两链表同时到达交点。此算法需O(n)的时间复杂度和O(1)的空间复杂度。

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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.

/**
 * 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 == NULL|| headB == NULL)
           return NULL;

        int lenA = 1, lenB = 1, diffe = 0;
        ListNode *A = headA;
        ListNode *B = headB;
        ListNode *Long = NULL;
        ListNode *Short = NULL;

        while( A->next != NULL){
            lenA++;
            A = A->next;
        }
        while( B->next != NULL){
            lenB++;
            B = B->next;
        }
        if(lenA > lenB){
            Long = headA;
            Short = headB;
            diffe = lenA - lenB;
        }
        else{
            Long = headB;
            Short = headA;
            diffe = lenB - lenA;
        }

        for(int i = 0; i < diffe; i++)
            Long = Long->next;

        while(Long != Short){
            Long = Long->next;
            Short = Short->next;
            if(Short == NULL)
              return NULL;
        }
        return Short;
    }
};
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