37 两个链表的第一个公共结点

输入两个链表，找出它们的第一个公共结点。

C++:

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* headA, ListNode* headB) {
        ListNode* p1 = headA ;
        ListNode* p2 = headB ;
        if (p1 == NULL || p2 == NULL)
            return NULL ;
        while(p1 != p2){
            if (p1 == NULL)
                p1 = headB ;
            else
                p1 = p1->next ;
            if (p2 == NULL)
                p2 = headA ;
            else
                p2 = p2->next ;
        }
        return p1 ;
    }
};

遍历得到两个链表的长度，求出他们之差，用的长的链表先走若干步，接着在同时在两个链表 上遍历，找到的第一个相同的结点就是他们的共同的结点

C++:

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    int GetListLen(ListNode* pHead){
        int len = 0 ;
        ListNode * pNode = pHead ;
        while(pNode != NULL){
            len++ ;
            pNode = pNode->next ;
        }
        return len ;
    }
    
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        int len1 = GetListLen(pHead1) ;
        int len2 = GetListLen(pHead2) ;
        int lenDif = len1 - len2 ;
        ListNode * pHeadLong = pHead1 ;
        ListNode * pHeadShort = pHead2 ;
        if (len2 > len1){
            lenDif = len2 - len1 ;
            pHeadLong = pHead2 ;
            pHeadShort = pHead1 ;
        }
        for (int i = 0 ; i < lenDif ; i++){
            pHeadLong = pHeadLong->next ;
        }
        while(pHeadLong != pHeadShort && pHeadLong != NULL && pHeadShort != NULL){
            pHeadLong = pHeadLong->next ;
            pHeadShort = pHeadShort->next ;
        }
        return pHeadLong ;
    }
};