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【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

52 兩個鏈表的第一個公共結點

題目描述

輸入兩個鏈表，找出它們的第一個公共結點。 /* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } };*/ #include <cmath> class Solution { public: ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) { //求兩鏈表的長度 int len1 = findListLength(pHead1); int len2 = findListLength(pHead2); ListNode* plong = pHead1, *pshort = pHead2; if(len1 < len2) { pshort = pHead1; plong = pHead2; } for(int i = 1; i <= abs(len1-len2); i++) plong = plong->next; //較長的鏈表多走幾步 //同時步進，直到遇到相同結點或者均遇到尾結點 while(plong != pshort) { plong = plong->next; pshort = pshort->next; } return plong; } private: int findListLength(ListNode* p) { int n = 0; while(p != nullptr) { p = p->next; n++; } return n; } }; Intersection of Two Linked Lists Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 begin to intersect at node c1. Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Credits: Special thanks to @stellari for adding this problem and creating all test cases.

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C++ //問題：求兩鏈表的交匯點 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /* 方法：雙指針法 (1) 如果有交匯點，p1掃描A，p2掃描B，掃描到結尾後，p1重定向到headB,p2重定向到headA,之後一定會在交匯點處相遇 因為交匯點之後都是路徑相同的，交匯點之前的路徑差可以由互換的兩次掃描中抵消 (2) 如果沒有交匯點，p1最後會到b末尾，p2會到a末尾，p1=p2=null，退出程序 O(m+n),O(1) */ class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *p1 = headA; ListNode *p2 = headB; if(p1 == NULL || p2 == NULL) return NULL;
while(p1 && p2 && p1!=p2) //只要不為空，進行掃描(註意，加上p1!=p2的判斷，可能兩鏈表長度為1，且相交，不加的話會返回null) { p1 = p1->next; p2 = p2->next; if(p1 == p2) return p1; //p1,p2同時為nullptr或者指向交匯點 if(p1 == NULL) p1 = headB; //重定向到另一個鏈表首結點 if(p2 == NULL) p2 = headA; } return p1; } };

【LeetCode & 劍指offer刷題】鏈表題5：52 兩個鏈表的第一個公共結點（Intersection of Two Linked Lists）