这题的难点在于理解单链表中如果有第一个公共节点，那么其后的所有节点都重合。
/* public class ListNode { int val; ListNode next = null; ListNode(int val) { this.val = val; } }*/ public class Solution { public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) { if(pHead1 == null || pHead2 == null){ return null; } //记录它们的长度可以不用两个辅助栈，让长的先走，可以同时到链尾 int nLen1 = GetListLength(pHead1); int nLen2 = GetListLength(pHead2); int nLenDif = nLen1 - nLen2; //不妨假定pHead1较长 ListNode pListHeadLong = pHead1; ListNode pListHeadShort = pHead2; if(nLen2 > nLen2){ pListHeadLong = pHead2; pListHeadShort = pHead1; nLenDif = nLen2 - nLen1; } //先在长链表上走几步，再同时在两个链表上遍历 for(int i = 0; i < nLenDif; i ++){ pListHeadLong = pListHeadLong.next; } while((pListHeadLong != null) && (pListHeadShort != null) && (pListHeadLong != pListHeadShort)){ pListHeadLong = pListHeadLong.next; pListHeadShort = pListHeadShort.next; } //得到第一个公共结点 return pListHeadLong; } private int GetListLength(ListNode pHead) { int nLen = 0; ListNode pNode = pHead; while(pNode != null){ ++ nLen; pNode = pNode.next; } return nLen; } }