Leetcode ☞ 160. Intersection of Two Linked Lists ☆

160. Intersection of Two Linked Lists

My Submissions

Question

Total Accepted: 64838  Total Submissions: 215467  Difficulty: Easy

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.

我的AC（32ms，落后2%的人）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if (!headA ||!headB)
        return NULL;
    struct ListNode *listA = headA;
    struct ListNode *listB = headB;
    struct ListNode *longerOne, *shorterOne;
    int lenA = 0, lenB = 0, sub;
    
    while(listA){
        lenA++;
        listA = listA->next;
    }
    while(listB){
        lenB++;
        listB = listB->next;
    }
    if (lenA > lenB){
        sub = lenA - lenB;
        longerOne = headA;
        shorterOne = headB;
    }
    else{
        sub = lenB - lenA;
        longerOne = headB;
        shorterOne = headA;
    }
    
    while(sub--){
        longerOne = longerOne->next;
    }
    while(longerOne){
        if (longerOne == shorterOne){
            return longerOne;
        }
        longerOne = longerOne->next;
        shorterOne = shorterOne->next;
    }
    return NULL;
}

思路：

分别遍历AB，计算出各自长度lenA，lenB。

然后再遍历，让长的链表从第（长len-短len）个节点开始，短的就从其head开始，当两个节点相同时，即为所求。没遇到说明没相交。

找环入口法（40ms）： Leetcode ☞ 142. Linked List Cycle II ☆

把A的尾连在B的头！【A尾连在A头，无意义】弄完记得还原！

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if (!headA || !headB) return NULL;
    
    struct ListNode *tailA = headA;
    while(tailA->next){
        tailA = tailA->next;
    }
    tailA->next = headB;
    //！！！！！！不能连到headA上！连到headA无意义！！！如果AB确实相交，那么就会成为一个环，否则，就成了一个新的长链表而已。
    
    struct ListNode *fast, *slow;
    fast = slow = headA;//链表A找环入口
    while(fast && fast->next){
        fast = fast->next->next;
        slow = slow->next;
        if (fast == slow)
            break;
    }
    if(!fast || !fast->next){
        tailA->next = NULL;
        return NULL;
    }
    
    fast = headA;
    while(fast != slow){
        fast = fast->next;
        slow = slow->next;
    }
    tailA->next = NULL;
    return fast;
}