160 Intersection of Two Linked Lists 相交链表

编写一个程序,找到两个单链表相交的起始节点。
例如,下面的两个链表:
A:           a1 → a2
                           
                                c1 → c2 → c3
                                        
B:  b1 → b2 → b3
在节点 c1 开始相交。
注意:
    如果两个链表没有交点,返回 null.
    在返回结果后,两个链表仍须保持原有的结构。
    可假定整个链表结构中没有循环。
    程序尽量满足 O(n) 时间复杂度,且仅用 O(1) 内存。
详见:https://leetcode.com/problems/intersection-of-two-linked-lists/description/

方法一:

/**
 * 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) {
        if(headA==nullptr||headB==nullptr)
        {
            return nullptr;
        }
        stack<ListNode*> stk1;
        stack<ListNode*> stk2;
        while(headA)
        {
            stk1.push(headA);
            headA=headA->next;
        }
        while(headB)
        {
            stk2.push(headB);
            headB=headB->next;
        }
        if(stk1.top()!=stk2.top())
        {
            return nullptr;
        }
        ListNode *common=nullptr;
        while(!stk1.empty()&&!stk2.empty()&&stk1.top()==stk2.top())
        {
            stk1.pop();
            common=stk2.top();
            stk2.pop();
        }
        return common;
    }
};

方法二:

/**
 * 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) {
        if(headA==nullptr||headB==nullptr)
        {
            return nullptr;
        }
        ListNode* head1=headA;
        ListNode* head2=headB;
        int n=0;
        while(head1)
        {
            ++n;
            head1=head1->next;
        }
        while(head2)
        {
            --n;
            head2=head2->next;
        }
        if(head1!=head2)
        {
            return nullptr;
        }
        head1=n>0?headA:headB;
        head2=head1==headA?headB:headA;
        n=abs(n);
        for(int i=0;i<n;++i)
        {
            head1=head1->next;
        }
        while(head1!=head2)
        {
            head1=head1->next;
            head2=head2->next;
        }
        return head1;
    }
};

原文地址:https://www.cnblogs.com/xidian2014/p/8728254.html

时间: 2024-10-25 11:38:00

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