7.两个排序链表的交点（Leetcode 160）

7.两个排序链表的交点（Leetcode 160）

方法一：

#include <stdio.h>

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

#include <set>

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
   		std::set<ListNode*> node_set;
		while(headA){
			node_set.insert(headA);
			headA = headA->next;
        }
        while(headB){
        	if (node_set.find(headB) != node_set.end()){
	        	return headB;
	        }
	        headB = headB->next;
        }
        return NULL;
    }
};

int main(){
	ListNode a1(1);
	ListNode a2(2);
	ListNode b1(3);
	ListNode b2(4);
	ListNode b3(5);
	ListNode c1(6);
	ListNode c2(7);
	ListNode c3(8);
	a1.next = &a2;
	a2.next = &c1;
	c1.next = &c2;
	c2.next = &c3;
	b1.next = &b2;
	b2.next = &b3;
	b3.next = &c1;
	
	Solution solve;
	ListNode *result = solve.getIntersectionNode(&a1, &b1);
	printf("%d\n", result->val);
	return 0;
}

方法二：

#include <stdio.h>

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

int get_list_length(ListNode *head){
	int len = 0;
	while(head){
		len++;
		head = head->next;
	}
	return len;
}

ListNode *forward_long_list(int long_len, 
				int short_len, ListNode *head){
	int delta = long_len - short_len;
	while(head && delta){
		head = head->next;
		delta--;
	}
	return head;
}

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
   		int list_A_len = get_list_length(headA);
   		int list_B_len = get_list_length(headB);   		
   		if (list_A_len > list_B_len){
   			headA = forward_long_list(list_A_len, list_B_len, headA);
	   	}
	   	else{
	   		headB = forward_long_list(list_B_len, list_A_len, headB);
	   	}   		
        while(headA && headB){
        	if (headA == headB){
	        	return headA;
	        }
	        headA = headA->next;
	        headB = headB->next;
        }
        return NULL;
    }
};

int main(){
	ListNode a1(1);
	ListNode a2(2);
	ListNode b1(3);
	ListNode b2(4);
	ListNode b3(5);
	ListNode c1(6);
	ListNode c2(7);
	ListNode c3(8);
	a1.next = &a2;
	a2.next = &c1;
	c1.next = &c2;
	c2.next = &c3;
	b1.next = &b2;
	b2.next = &b3;
	b3.next = &c1;	
	Solution solve;
	ListNode *result = solve.getIntersectionNode(&a1, &b1);
	printf("%d\n", result->val);
	return 0;
}