leetCode 160. Intersection of Two Linked Lists 鏈表

160. 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.

題目大意：

找出兩個鏈表後半部分的交匯點。

思路：

1.求出兩個鏈表的長度。

2.獲取鏈表長度差n。

3.將長的鏈表先移動到第n個節點。

4.對長鏈表和短鏈表進行比較。(同時向後移動)如果在鏈表尾之前找到相等的節點，返回該節點，如果沒找到，返回NULL。

代碼如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int listLength(ListNode *head)//用快指針求鏈表長度
    {
        ListNode * p = head;
        int i = 0 ;
        while(p && p->next)
        {
            i++;
            p = p->next->next;
        }
        if(p == NULL)
            return 2 * i;
        return 2 * i + 1;
    }
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int lenA = listLength(headA);
        int lenB = listLength(headB);
        
        int maxLen = lenA > lenB ? lenA :lenB;
        int remain ; 
        ListNode * la,*lb;
        la = headA;
        lb = headB;
        
        if(maxLen == lenA)
        {
            remain = lenA - lenB;
            while(remain--)
            {
                la = la->next;
            }
        }
        else
        {
            remain = lenB - lenA;
            while(remain--)
                lb = lb->next;
        }
        
        while(lb != NULL)
        {
            if(la != lb)
            {
                la = la->next;
                lb = lb->next;
            }
            else
            {
                return la;
            }
        }
        return NULL;
    }
};

2016-08-13 01:08:14