这篇文章主要讲解了“SpringBoot怎么找出两个单链表的交叉节点”,文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习“SpringBoot怎么找出两个单链表的交叉节点”吧!
题目:写一个程序找出两个单链表的交叉节点。
思路:单链表A和单链表B,交叉点后的部分是一样的,也就是说长度是一样的,如上所示:c1 → c2 → c3。所以,将单链表A和单链表B相差的部分去掉,依次对应比较等长的部分即可。在计算两个链表的长度之后,比较两个链表的尾节点是否一样,如果不一样说明没有交叉节点,返回NULL。
Language : c
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {struct ListNode *curA = (struct ListNode*)malloc(sizeof(struct ListNode));struct ListNode *curB = (struct ListNode*)malloc(sizeof(struct ListNode)); curA = headA; curB = headB;int length_a = 1;int length_b = 1;int i = 0;if(curA == NULL || curB == NULL){return NULL; }while(curA->next != NULL){ curA = curA->next; length_a++; }while(curB->next != NULL){ curB = curB->next; length_b++; }if(curA != curB){return NULL; } curA = headA; curB = headB;if(length_a > length_b){for(i; i < length_a-length_b; i++){ curA = curA->next; } i = 0; }else if(length_a < length_b){for(i; i < length_b-length_a; i++){ curB = curB->next; } i = 0; }while(curA != curB){ curA = curA->next; curB = curB->next; }return curA; }
Language : cpp
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *curA, *curB; curA = headA; curB = headB;if(curA == NULL || curB == NULL){return NULL; }int length_a = getLength(curA);int length_b = getLength(curB);if(length_a > length_b){for(int i=0; i < length_a-length_b; i++){ curA = curA->next; } }else if(length_a < length_b){for(int i=0; i < length_b-length_a; i++){ curB = curB->next; } }while(curA != curB){ curA = curA->next; curB = curB->next; }return curA; } private:int getLength(ListNode *head){int length = 1;while(head->next != NULL){ head = head->next;length++; }return length; } };
Language:python
# Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = Noneclass Solution(object):def getIntersectionNode(self, headA, headB):""" :type head1, head1: ListNode :rtype: ListNode """if headA is None or headB is None:return Nonepa = headA # 2 pointerspb = headBwhile pa is not pb:# pa先遍历headA,然后再遍历headB# pb先遍历headB,然后再遍历headApa = headB if pa is None else pa.next pb = headA if pb is None else pb.nextreturn pa # 只有两种方式结束循环,一种是pa和pb所指相同,另一种是headA和headB都已经遍历完仍然没有找到。
感谢各位的阅读,以上就是“SpringBoot怎么找出两个单链表的交叉节点”的内容了,经过本文的学习后,相信大家对SpringBoot怎么找出两个单链表的交叉节点这一问题有了更深刻的体会,具体使用情况还需要大家实践验证。这里是亿速云,小编将为大家推送更多相关知识点的文章,欢迎关注!
原创文章,作者:carmelaweatherly,如若转载,请注明出处:https://blog.ytso.com/tech/opensource/220252.html