SpringBoot怎么找出两个单链表的交叉节点

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，小编将为大家推送更多相关知识点的文章，欢迎关注！