剑指offer——(10)两个链表的第一个公共结点

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图解: 

代码: 

public class Solution {
/*
    遍历两条链表得到各自的长度,长的先next两条链表的长度差,然
    后一起走next,如果有公共结点,此结点及之后的链表值都相同。
*/
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
        if(pHead1==null||pHead2==null) return null;        
        ListNode pre1 = pHead1,pre2 = pHead2;
        while(pre1!=null){
           length1++;
           pre1 = pre1.next;
        }
        while(pre2!=null){
            length2++;
            pre2 = pre2.next;
        }
        if(length1>=length2) {
            for(int i=1;i<=length1-length2;i++){
                pHead1 = pHead1.next;
            }
        }
        else{
            for(int i=1;i<=length2-length1;i++)
                pHead2 = pHead2.next;           
        }
        while(pHead1!=pHead2){
            pHead1 = pHead1.next;
            pHead2 = pHead2.next;  
        }
        return pHead1;     
    }
}

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/

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转载自blog.csdn.net/qq_38071429/article/details/83956399