《leetCode》：Intersection of Two Linked Lists

题目

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.

思路

struct ListNode {
int val;
struct ListNode *next;
};
int getListLen(struct ListNode *head){
if(head==NULL){
return 0;
}
struct ListNode *cur=head;
int len=0;
while(cur!=NULL){
len++;
cur=cur->next;
}
return len;
}
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
if(headA==NULL||headB==NULL){
return NULL;
}
//计算出链表A、B的长度差
int listALen=getListLen(headA);
int listBLen=getListLen(headB);
struct ListNode *curLargeList=NULL;
struct ListNode *curLowList=NULL;
int dif=0;
if(listALen>=listBLen){//注意：等号如果不写，下面这个if要写
dif=listALen-listBLen;
curLargeList=headA;
curLowList=headB;
}
if(listALen<listBLen){
dif=listBLen-listALen;
curLargeList=headB;
curLowList=headA;
}
while(dif>0){
curLargeList=curLargeList->next;
dif--;
}
while(curLargeList!=curLowList){
curLargeList=curLargeList->next;
curLowList=curLowList->next;
}
return curLargeList;
}