36两个链表的第一个公共结点

 

题目描述

输入两个链表，找出它们的第一个公共结点。

思路1：
首次遍历:算出2个链表的长度l1,l2。
第二次遍历,长的链表先走|l2-l1|步，然后2个链表同时遍历，找到第一个相同的节点输出。

public class Solution {
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
         int l1 = 0,l2 = 0,ldiff=0;
        ListNode longp,shortp;
        
        // 得到长度
        for(ListNode p = pHead1;p!=null;p=p.next)
            l1++;
        for(ListNode p = pHead2;p!=null;p=p.next)
            l2++;
        if(l1>l2){
            ldiff = l1-l2;
            longp=pHead1;shortp = pHead2;
        }
           else{
             ldiff = l2-l1;
            longp=pHead2;shortp = pHead1;
        }
        //长链表先走
        for(int i = 0;i<ldiff;i++){
            longp = longp.next;
        }
        //2个链表同时遍历
        while(longp!=null){
            if(longp.val==shortp.val) return longp;
            longp = longp.next;
            shortp = shortp.next;
            
        }
        return null; 
    }
}

C++:20180726

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        //find coross
        ListNode* fast = pHead1;
        ListNode* slow = pHead2;
        int l1 = 0,l2 = 0,d=0;
        while(fast!=NULL) {fast=fast->next;l1++;}
        while(slow!=NULL) {slow=slow->next;l2++;}
        if(l1>l2) {
            fast = pHead1;
            slow = pHead2;
            d = l1-l2;
        }
        else{
            fast = pHead2;
            slow = pHead1;
            d = l2-l1;
        } 
        
        for(int i = 0;i<d;i++)
            fast = fast->next;
        while(fast!=NULL){
            if(fast==slow)
                    return fast;
            fast = fast->next;
            slow = slow->next;
        }
        return NULL;
    }
};