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LeetCode 160 :Intersection of Two Linked Lists

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标签:leetcode   intersection of two   链表交点   c++   

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.
分析:
找出两个链表的交点,有两种方法:
1)暴力法,
将链表A中的所有元素加入到Map中。
对链表B中得每一个元素,判断是不是在Map中,如果在,则是A和B的交叉点。如果都不是,则A和B没有交叉点。
该方法肯定不能满足题目中要求的O(n)和O(1)的时间、空间复杂度要求。
2)计算长度法。假设链表A和B相交,则交点及之后的部分长度是相同的,相差的是交点前的部分。计算A和B的长度差n,较长的那个先走N步,然后和较短的那个同时遍历。如果指针指向同一元素,则为交点,否则A和B 没有交点。该方法能够满足线性时间复杂度和常数空间复杂度的要求。
代码如下:
class Solution {
public:

    //计算链表长度
  int calcListLength(ListNode *head){
        int count = 0;
        while (head) {
            count++;
            head = head->next;
        }
        return count;
    }
    
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int lengthA = calcListLength(headA);
        int lengthB = calcListLength(headB);
        ListNode *pA = headA, *pB = headB;
        //计算链表长度的差值,较长的先走N步 
        int n = lengthA - lengthB;
        for(int i=0; i<abs(n); i++){
            if(n>0){
                pA = pA->next;
            }else{
                pB = pB ->next;
            }
        }
        //同时开始遍历,如相等则为交叉点
        while (pB && pA) {
            if (pB == pA) {
                return pB;
            }
            pB = pB->next;
            pA = pA ->next;
        }
        return NULL;
        
    }
};



LeetCode 160 :Intersection of Two Linked Lists

标签:leetcode   intersection of two   链表交点   c++   

原文地址:http://blog.csdn.net/sunao2002002/article/details/45727149

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