LeetCode 160. Intersection of Two Linked Lists

原题链接在这里：https://leetcode.com/problems/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.

题解：

找到length diff, 长的list head先移动diff次, 再一起移动找相同点.

Time Complexity: O(len1 + len2), len1 is the length of list one. len2 is the length of list two.

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int len1 = length(headA);
        int len2 = length(headB);
        while(len1 > len2){
            headA = headA.next;
            len1--;
        }
        
        while(len2 > len1){
            headB =headB.next;
            len2--;
        }
        
        while(headA != headB){
            headA = headA.next;
            headB = headB.next;
        }
        
        return headA;
    }
    
    private int length(ListNode head){
        int len = 0;
        while(head != null){
            head = head.next;
            len++;
        }
        return len;
    }
}

有一巧妙地方法来综合掉 length diff, a = headA, b = headB, a和b一起移动。当a到了list A的末位就跳到HeadB, b到了List B的末位就跳到HeadA.

等a和b相遇就是first intersection node. 因为a把first intersection node之前的list A部分, list B部分都走了一次. b也是如此. diff就综合掉了.

若是没有intersection, 那么a走到list B的结尾 null时, b正好走到 list A的结尾null, a==b. 返回了null.

Time Complexity: O(len1 + len2), len1 is the length of list one. len2 is the length of list two.

Space: O(1).

AC  Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode a = headA;
        ListNode b = headB;
        while(a != b){
            a = a==null ? headB : a.next;
            b = b==null ? headA : b.next;
        }
        return a;
    }
}

AC C++:

/**
 * 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) {
        ListNode* p1 = headA;
        ListNode* p2 = headB;
        while(p1 != p2){
            p1 = p1 ? p1->next : headB;
            p2 = p2 ? p2->next : headA;
        }

        return p1;
    }
};

AC Python:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        p1 = headA
        p2 = headB
        while p1 is not p2:
            p1 = p1.next if p1 else headB
            p2 = p2.next if p2 else headA
        return p1