leetcode | Intersection of Two Linked Lists

Intersection of Two Linked Lists : https://leetcode.com/problems/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:



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.

解析：

我们从例程中可以看出，如果两个链表中的所有节点两两对比，则时间复杂度O(n2)O(n^2)

如果在两个链表相遇之前的两个链表长度如果是相等的，那么，我们只需上下比较2个链表的对应元素，即可找到是否存在的交叉点。因此，可以首先统计2个链表的长度lenA, lenB，让长度更长的链表先走，如A更长，则A先走(lenA-lenB)步，然后A,B一块走，同时比较A,B对应的节点是否相同。时间复杂度O(m+n)=O(n)

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/

//Intersection 交叉点，注意O(n)可以遍历链表多次
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL)
return NULL;
int lenA = getListLen(headA);
int lenB = getListLen(headB);
if (lenA > lenB) {
while (lenA != lenB) {
headA = headA->next;
lenA--;
}
} else {
while (lenA != lenB) {
headB = headB->next;
lenB--;
}
}
while (headA != headB) {
headA = headA->next;
headB = headB->next;
}
return headA;
}
private:
int getListLen(ListNode* head) {
int len = 0;
while (head) {
len++;
head = head->next;
}
return len;
}
};