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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> LeetCode—Intersection of Two Linked Lists 找兩個鏈表的相交位置,讓長的鏈表先走一段

LeetCode—Intersection of Two Linked Lists 找兩個鏈表的相交位置,讓長的鏈表先走一段

編輯:C++入門知識

LeetCode—Intersection of Two Linked Lists 找兩個鏈表的相交位置,讓長的鏈表先走一段


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. 只能夠是O(n)的時間復雜度,這裡的做法是首先遍歷兩個鏈表知道長度,然後讓長的鏈表先走一段,然後再進行兩兩比較,時間復雜度O(lengthA+lengthB+maxLength)就是O(n)

     

    /**
     * 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 == NULL || headB == NULL)
            {
                return NULL;
            }
            int lengthA = 1;
            int lengthB = 1;
            ListNode *Afirst = headA;
            ListNode *Bfirst = headB;
            while(Afirst->next)
            {
                lengthA++;
                Afirst = Afirst->next;
            }
            while(Bfirst->next)
            {
                lengthB++;
                Bfirst = Bfirst->next;
            }
            if(Afirst != Bfirst)
            {
                return NULL;
            }
            int length = lengthA;//<隨便初始化一個
            if(lengthA > lengthB)
            {
                int diff = lengthA-lengthB;
                length = lengthB;
                while(diff--)
                {
                    headA = headA->next;
                }
            }
            if(lengthA < lengthB)
            {
                length = lengthA;
                int diff = lengthB-lengthA;
                while(diff--)
                {
                    headB = headB->next;
                }
            }
            while(length--)
            {
                if(headA == headB)
                {
                    return headA;
                }
                headA = headA->next;
                headB = headB->next;
            }
            return NULL;
            
        }
    };


     

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