2015-10-18 10:42:52 +08:00

75 lines
2.2 KiB
C++

// Source : https://oj.leetcode.com/problems/intersection-of-two-linked-lists/
// Author : Hao Chen
// Date : 2014-12-01
/**********************************************************************************
*
* 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) {
//caculate the length of each List
int lenA = getListLength(headA);
int lenB = getListLength(headB);
if (lenA<=0 || lenB<=0 ) return NULL;
//let List A is the longest List;
if (lenA < lenB){
swap(headA, headB);
}
//move head of List A, make both of Lists are same length
for (int i=0; i<abs(lenA-lenB); i++){
headA = headA->next;
}
//synced travel both of Lists and check their nodes are same or not
while (headA != headB){
headA = headA->next;
headB = headB->next;
}
return headA;
}
private:
inline int getListLength(ListNode *head){
int len=0;
while(head!=NULL){
head = head->next;
len++;
}
return len;
}
};