[劍指offer專題]兩個鏈表的第一個公共結點

//方法一：運用HashMap的特性
import java.util.HashMap;

public class Solution {
    public ListNode findFristCommonNode(ListNode pHead1,ListNode pHead2){
        ListNode current1 = pHead1;
        ListNode current2 = pHead2;
        Map<ListNode,Integer> haspMap = new HashMap<ListNode,Integer>();
        while(current1 != null){
            hashMap.put(currnet1,null);
            current1 = current1.next;
        }
        while(current2 != null){
            if(haspMap.comtainsKey(current2)){
                return current2;
            }
        }
        
        return null;  
    }
}

//方法2 長鏈表先走 len1 － len2
public ListNode findFirstCommonNode(ListNode pHead1,ListNode pHead2){
    ListNode current1 = pHead1;
    ListNode current2 = pHead2;
    if(pHead1 == null || pHead2 == null){
        return null;
    }
    
    int len1 = getLength(pHead1);
    int len2 = getLength(pHead2);
    
    if(len1 > len2){
        int len = len1 - len2;
        while(len>0){
            current1 = current1.next;
            len--;
        }
    }else if(len1 < len2){
        int len = len2 - len1;
        while(len>0){
            currnet2 = current2.next;
            len--;
        }
    }
    
    while(current1 != null && current2 != null){
        if(current1 == current2){
            return current1;
        }
        current1 = currnet1.next;
        current2 = current2.next;
    }
    return null;
}

public int getLength(ListNode pHead){
    ListNode current = pHead;
    int count=0;
    while(current != null){
        count++;
        current = current.next;
    }
    return count;
}