leetcode 160: Intersect Linked Lists

Subject: Intersecting Linked Lists

• Topic description:
  Write a program to find the starting node where two singly linked lists intersect.

For example, the following two linked lists:

A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3

The intersection starts at node c1.
Notice:

• Returns null if the two linked lists do not intersect.
• After returning the result, the two linked lists must still maintain the original structure.
• It can be assumed that there are no loops in the entire linked list structure.
• The program tries to satisfy O(n) time complexity and only uses O(1) memory.

Solution one:

[]
/**
 * 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) {
        std::set<ListNode*> node_set;  //设置查找集合node_set
        while(headA){
            node_set.insert(headA);  //将链表A中的节点插入node_set
            headA = headA->next;  //遍历链表A
        }
        while(headB){
            if(node_set.find(headB)!=node_set.end()){
                return headB;  //当在headB中找到第一个出现在node_set中的节点时
            }
            headB = headB->next;   //遍历链表B
        }
        return NULL;
    }
};

Solution two:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
int get_list_length(ListNode *head){
    int len = 0;
    while(head){  //遍历链表，计算链表长度
        len++;
        head = head->next;
    }
    return len;
}

ListNode *forward_long_list(int long_len,int short_len,ListNode *head){
    int delta = long_len-short_len;
    while(head && delta){  //将指针向前移动至多出节点个数后面的位置
        head = head->next;
        delta--;
    }
    return head;
}
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int list_A_len = get_list_length(headA);
        int list_B_len = get_list_length(headB);
        if(list_A_len>list_B_len){  //如果链表A长，移动headA到对应位置
            headA = forward_long_list(list_A_len,list_B_len,headA);
        }else{  //如果链表B长，移动headB到对应位置
            headB = forward_long_list(list_B_len,list_A_len,headB);
        }
        while(headA && headB){
            if(headA==headB){   //当两个指针指向了同一个节点时，说明找到了
                return headA;
            }
            headA = headA->next;
            headB = headB->next;
        }
        return NULL;
    }
};