《剑指offer》-两个链表的第一个公共结点

题目描述

输入两个链表，找出它们的第一个公共结点。

解法：

一、暴力法，第一个链表从开头遍历，和第二个链表的每个节点元素比较，则需要两个for嵌套循环，那么复杂度是o（mn）

二、低复杂度法，最长的链表先行遍历，然后再同时遍历比较相等的节点。复杂度为o(n+m)

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