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.

解題思路:

1、可以將A,B兩個鏈表看做兩部分,交叉前與交叉後。

2、交叉後的長度是一樣的,因此交叉前的長度差即爲總長度差。

3、只要去除這些長度差,距離交叉點就等距了。

4、爲了節省計算,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;
        int lenA = 1;
        int lenB = 1;
        ListNode *tempA = headA;
        ListNode *tempB = headB;
        
        while(tempA->next != NULL){
            tempA = tempA->next;
            lenA++;
        }
        while(tempB->next != NULL){
            tempB = tempB->next;
            lenB++;
        }
        
        if(tempA != tempB)
            return NULL;
        if(lenA>lenB){
            for(int i=0;i<lenA-lenB;i++){
                headA = headA->next;
            }
        }
        if(lenB>lenA){
            for(int i=0;i<lenB-lenA;i++){
                headB = headB->next;
            }
        }
        while(headA != headB){
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
    }
};



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