上周面試掛了，反思原因，莫非是因?yàn)橐坏浪惴}沒做好嗎？這題目是“判斷兩條鏈表是否交叉，若有交叉，返回交叉節(jié)點(diǎn)的指針。” 為了防止反復(fù)在同一個(gè)陰溝里翻船，決定把最優(yōu)解寫出來。

#include "pch.h" #include <iostream> #include<assert.h>template<typename T> class List { public:struct Node {T data;Node* next = nullptr;Node(T& d) : data(d) {}};Node* m_head = nullptr; private:enum {E_SUCCESS = 0,E_FAIL = -1}; public://List(){}~List() {Node* pcur = m_head, *pnext = nullptr;while (pcur) {pnext = pcur->next;delete pcur;pcur = pnext;}}int AddNode(T dt, bool at_tail = true) {Node*ptr = new Node(dt);return AddNode(ptr, at_tail);}int AddNode(Node* nd, bool at_tail = true) {if (nd) {if (at_tail) {Node* ptail = Tail();if (ptail) {ptail->next = nd;}else {m_head = nd;}}else {nd->next = m_head;m_head = nd;}return E_SUCCESS;}return E_FAIL;}Node* Intersect(List& oth) {Node* entrance1 = LoopEntrance();Node* entrance2 = oth.LoopEntrance();if (entrance1 == entrance2) {return entrance1;}else if (!entrance1 && !entrance2) {Node* longlist_ptr = nullptr, *shortlist_ptr = nullptr;{const int sz1 = Size();const int sz2 = oth.Size();int num = 0;if (sz1 >= sz2) {longlist_ptr = m_head;shortlist_ptr = oth.m_head;num = sz1 - sz2;}else {longlist_ptr = oth.m_head;shortlist_ptr = m_head;num = sz2 - sz1;}for (int i = 0; i < num; ++i) {if (longlist_ptr) {longlist_ptr = longlist_ptr->next;}}}while (longlist_ptr && shortlist_ptr) {if (longlist_ptr == shortlist_ptr) {return longlist_ptr;}longlist_ptr = longlist_ptr->next;shortlist_ptr = shortlist_ptr->next;}}return nullptr;}Node* LoopEntrance() {Node *fast_ptr = m_head, *slow_ptr = m_head;bool has_loop = false;while (fast_ptr && fast_ptr->next) {fast_ptr = fast_ptr->next->next;slow_ptr = slow_ptr->next;if (fast_ptr == slow_ptr) {has_loop = true;break;}}if (has_loop) {slow_ptr = m_head;while (1 /*fast_ptr && fast_ptr->next && slow_ptr*/) {if (fast_ptr == slow_ptr) {return fast_ptr;}fast_ptr = fast_ptr->next;slow_ptr = slow_ptr->next;}}return nullptr;}private:Node* Tail() {if (LoopEntrance()) {return nullptr;}Node* ptr = m_head;while (ptr && ptr->next) {ptr = ptr->next;}return ptr;}int Size() {int sz = 0;Node* ptr = m_head;Node* entrance = LoopEntrance();if (entrance) {while (ptr && ptr != entrance) {sz++;ptr = ptr->next;}assert(ptr == entrance);sz++; // entrance pointwhile (ptr && ptr->next != entrance) {sz++;ptr = ptr->next;}}else {while (ptr) {sz++;ptr = ptr->next;}}return sz;} };int main() {std::cout << "Hello World!\n"; List<int> list1, list2;for (int i = 1; i < 3; ++i) {list1.AddNode(i);}for (int i = 25; i < 28; ++i) {list2.AddNode(i);}#if 1for (int i = 100; i < 101; ++i) {List<int>::Node* node = new List<int>::Node(i);list1.AddNode(node);list2.AddNode(node);} #endifauto ptr = list1.Intersect(list2);if (ptr) {std::cout << "intersect point at " << ptr << std::endl;}else {std::cout << "no intersect\n";}return 0; }

代碼創(chuàng)建了兩條有交叉節(jié)點(diǎn)的鏈表。如圖所示：

程序運(yùn)行結(jié)束時(shí)，析構(gòu)這兩條鏈表，發(fā)生錯(cuò)誤，是因?yàn)榻徊婀?jié)點(diǎn)被兩次釋放：

轉(zhuǎn)載于:https://blog.51cto.com/frankniefaquan/2378379

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