目錄
鏈表題目：
1.從尾到頭打印鏈表
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* ListNode(int x) :
* val(x), next(NULL) {
* }
* };
*/
class Solution {
public:
vector<int> printListFromTailToHead(ListNode* head) {
// 使用棧來保存
if (head == nullptr) return vector<int>();
stack<int> mystack;
vector<int> res;
ListNode* p = head;
while (p != nullptr){
mystack.push(p->val);
p = p->next;
}
while (!mystack.empty()){
res.push_back(mystack.top());
mystack.pop();
}
return res;
}
};
2.鏈表的倒數第k個結點
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
// 設置一個結點先走k步，然後一起走
if (pListHead == nullptr) return NULL;
if (k < 0) return NULL;
ListNode* p = pListHead;
for (int i = 0; i< k-1; i++){
p = p->next;
if (p == nullptr) return NULL;
}
while (p->next != nullptr){
p = p->next;
pListHead = pListHead->next;
}
return pListHead;
}
};
3. 翻轉鏈表
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* ReverseList(ListNode* pHead) {
if (pHead == nullptr || pHead-> next == nullptr) return pHead;
ListNode* p = pHead;
ListNode* newP = NULL;
ListNode* temp = NULL;
while (p){
temp = p->next;
p->next = newP;
newP = p;
p = temp;
}
return newP;
}
};
遞歸的形式翻轉鏈表：
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* ReverseList(ListNode* pHead) {
// 使用遞歸的方法來翻轉一個鏈表
if (pHead == nullptr || pHead->next == nullptr){
return pHead;
}
ListNode* reversedP;
reversedP = ReverseList(pHead->next);
pHead->next->next = pHead;
pHead->next = NULL;
return reversedP;
}
};
4.合併2 個有序的鏈表
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
// 使用遞歸的方法合併連個有序的鏈表
if (pHead1 == nullptr && pHead2 != nullptr){
return pHead2;
}
if (pHead2 == nullptr && pHead1 != nullptr){
return pHead1;
}
if (pHead1 == nullptr && pHead2 == nullptr){
return NULL;
}
ListNode* ret = NULL; // 先在外部進行聲明
if (pHead1->val >= pHead2-> val){
ret = pHead2;
ret->next = Merge(pHead1, pHead2->next);
}else{
ret = pHead1;
ret->next = Merge(pHead1->next, pHead2);
}
return ret;
}
};
5. 兩個鏈表的第一個公共結點
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
// 先計算2 個鏈表的長度，然後計算差值k，短的先走k步，然後一起走，知道有相同的結點
if (pHead1 == nullptr || pHead2 == nullptr){
return NULL;
}
int len1= 0;
int len2 = 0;
ListNode* move1 = pHead1;
ListNode* move2 = pHead2;
while (move1 != nullptr){
len1 ++;
move1 = move1->next;
}
while (move2 != nullptr){
len2++;
move2 = move2->next;
}
while (len1 > len2){
len1 --;
pHead1 = pHead1->next;
}
while (len1 < len2){
len2 --;
pHead2 = pHead2->next;
}
// 一起往前走
while (pHead1 != nullptr){
if (pHead1 == pHead2){
return pHead1;
}else{
pHead1 = pHead1->next;
pHead2 = pHead2->next;
}
}
}
};
6. 鏈表中環的入口
設置快慢結點，找到相遇的結點；然後首結點和相遇的節點一起往前走，直到找到相交的結點。
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* EntryNodeOfLoop(ListNode* pHead)
{
// 設置快慢結點，走1 步和走2步，找到相遇的結點q； 然後q 和頭部結點一起往前走，直到找到相遇的結點；
if (!pHead || !pHead->next || !pHead->next->next) return NULL;
ListNode* low = pHead->next;
ListNode* fast = pHead->next->next;
while (low != fast){
if (!low->next || !fast->next->next) return NULL;
low = low->next;
fast = fast->next->next;
}
while (fast != pHead){
pHead = pHead->next;
fast = fast->next;
}
return pHead;
}
};
7. 刪除鏈表中重複的節點
重複的結點全部刪除
1. 設置dummy 結點； 2. 設置一個 前置結點last 和當前結點 pHead， 使用last -> next 指向pHead ,即刪除結點
3. 設置初始的判斷條件是 while (pHead && pHead -> next)
創建一個結點： ListNode* node = new ListNode(-1);
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* deleteDuplication(ListNode* pHead)
{
if (!pHead || !pHead->next) return pHead;
// 創建一個dummy 結點
ListNode* first = new ListNode(-1);
first->next = pHead;
ListNode* last = first;
while(pHead && pHead->next){
if (pHead->val == pHead->next->val){
int values = pHead->val;
while (pHead && pHead->val == values){
pHead = pHead->next;
}
last->next = pHead;
}else{
last = last->next;
pHead = pHead->next;
}
}
return first->next;
}
};