找到两个单链表相交的起始节点。

https://leetcode-cn.com/problems/intersection-of-two-linked-lists/

#define _CRT_SECURE_NO_WARNINGS 
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//找到两个单链表相交的起始节点。https://leetcode-cn.com/problems/intersection-of-two-linked-lists/description/

typedef struct ListNode{
	char val[3];
	struct ListNode *next;
}Node;

//方法一：走两遍法
struct ListNode* getIntersectionNode(Node*headA,Node*headB){
	struct ListNode* curA = headA;
	struct ListNode* curB = headB;
	while (curA != curB){
		if (curA){
			curA = curA->next;
		}
		else{
			curA = headB;
		}
		if (curB){
			curB = curB->next;
		}
		else{
			curB = headA;
		}
	}
	return curB;
	
}
//方法二： 长链表先走差步 法；
struct ListNode* getIntersectionNode_2(Node*headA, Node*headB){
	Node*curA = headA;
	Node*curB = headB;
	//先计算链表A,B的长度
	int La = 0, Lb = 0;
	while (curA){
		++La;
		curA = curA->next;
	}
	while (curB){
		++Lb;
		curB = curB->next;
	}
	//找出长链表和短链表 ；（假设链表A长，B短；）
	Node*longList = headA;
	Node*shortList = headB;
	if (La < Lb){
		longList = headB;
		shortList = headA;
	}
	int gap = abs(La - Lb);
	while (gap){
		longList = longList->next;
		gap--;
	}
//============================================
	while (longList!= shortList){
		longList = longList->next;
		shortList = shortList->next;
	}
//----------上段也可以这么写-----------------
	/*while (longList){
		if (longList == shortList){
			return longList;
		}
		longList = longList->next;
		shortList = shortList->next;
	}
	return NULL;			*/
//=============================================
	return longList;
}
int main(){
	Node*a1 = (Node*)malloc(sizeof(Node));
	strcpy(a1->val, "a1");   //a1->val= "a1";  字符串赋值只能 “strcpy”
	Node*a2 = (Node*)malloc(sizeof(Node));
	strcpy(a2->val, "a2");

	Node*c1 = (Node*)malloc(sizeof(Node));
	strcpy(c1->val, "c1");
	Node*c2 = (Node*)malloc(sizeof(Node));
	strcpy(c2->val, "c2");
	Node*c3 = (Node*)malloc(sizeof(Node));
	strcpy(c3->val, "c3");

	Node*b1 = (Node*)malloc(sizeof(Node));
	strcpy(b1->val,"b1");
	Node*b2 = (Node*)malloc(sizeof(Node));
	strcpy(b2->val , "b2");

	//链表1：
	a1->next = a2; a2->next = c1; c1->next = c2; c2->next = c3; c3->next = NULL;
	//链表2：
	b1->next = b2; b2->next = c1; c1->next = c2; c2->next = c3; c3->next = NULL;

	//Node*ret = getIntersectionNode(a1, b1);
	Node*ret = getIntersectionNode_2(a1, b1);  //答案：c1
	printf("Reference of the node with value:  %s\n", ret->val);
	system("pause");
	return 0;
}