我的思路：先求长度，每个长度长的遍历一次长度短的，
后看评论理解题意：第一个公共结点后面的都相同，所以可以先走长-短的差再比较。
public class Solution {
public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
if(pHead1 == null || pHead2 ==null)
return null;
else if(pHead1 != null && pHead2!=null){
int i,j; //求长度
ListNode p1 = pHead1;
ListNode p2 = pHead2;
for(i=1; p1 != null; i++){
p1 = p1.next;
}
for(j=1; p2 != null; j++){
p2 = p2.next;
}
if(i>j){ //换位，减少代码量
ListNode tem = pHead1;
pHead1 = pHead2;
pHead2 = tem;
}
ListNode q1 = null;
while(pHead2 != null){
q1 = pHead1;
while(q1 != null && pHead2.val != q1.val) //注意判断的顺序，颠倒会发生nullpointerexception异常
{q1 = q1.next;}
if(q1 != null && pHead2.val == q1.val) return pHead2;
else pHead2 = pHead2.next;
}
}
return null;
}
}
简法：
public class Solution {
public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
if(pHead1 == null || pHead2 ==null)
return null;
else if(pHead1 != null && pHead2!=null){
int i,j; //求长度
ListNode p1 = pHead1;
ListNode p2 = pHead2;
for(i=1; p1 != null; i++){
p1 = p1.next;
}
for(j=1; p2 != null; j++){
p2 = p2.next;
}
if(i>j){ //换位，减少代码量
ListNode tem = pHead1;
pHead1 = pHead2;
pHead2 = tem;
}
for(int z=j-i; z>0; z--){
pHead2 = pHead2.next;}
while(pHead1 != pHead2){
pHead1 = pHead1.next;
pHead2 = pHead2.next;}
return pHead1;
}
return null;
}
}
附大神解法：
class Solution {public: ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) { ListNode *p1 = pHead1; ListNode *p2 = pHead2; while(p1!=p2){ p1 = (p1==NULL ? pHead2 : p1->next); p2 = (p2==NULL ? pHead1 : p2->next); } return p1; }};