【LeetCode】160. Intersection of Two Linked Lists

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.

 

可以將A,B兩個鏈表看做兩部分，交叉前與交叉后。

交叉后的長度是一樣的，因此交叉前的長度差即為總長度差。

只要去除這些長度差，距離交叉點就等距了。

為了節省計算，在計算鏈表長度的時候，順便比較一下兩個鏈表的尾節點是否一樣，

若不一樣，則不可能相交，直接可以返回NULL

/**
 * 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;
        ListNode* iter1 = headA;
        ListNode* iter2 = headB;
        int len1 = 1;
        while(iter1->next != NULL)
        {
            iter1 = iter1->next;
            len1 ++;
        }
        int len2 = 1;
        while(iter2->next != NULL)
        {
            iter2 = iter2->next;
            len2 ++;
        }
        if(iter1 != iter2)
            return NULL;
        if(len1 > len2)
        {
            for(int i = 0; i < len1-len2; i ++)
                headA = headA->next;
        }
        else if(len2 > len1)
        {
            for(int i = 0; i < len2-len1; i ++)
                headB = headB->next;
        }
        while(headA != headB)
        {
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
    }
};