今天研究了一下markdown的語法才發(fā)現(xiàn)還有一種可以劃分出區(qū)域的方法。
鏈表是一種很常見的數(shù)據(jù)結(jié)構(gòu)童漩，那么我們就復習一下悦荒，使用Ｃ++現(xiàn)擼出一個linkedlist

#include <iostream>
using namespace std;

// Node類,封裝了一些常用的操作
template <typename T>
class Node{
public:
    //構(gòu)造函數(shù)
    Node();
    Node(T data);
    //析構(gòu)函數(shù)
    ~Node();

    void setData(T data);
    T getData();
    void setNext(Node<T> *next);
    Node* getNext();
    void printData();
private:
    T* m_tpData;
    Node<T> *m_tpNext;
}

template <typename T>
// 不帶參數(shù)的構(gòu)造函數(shù)
Node<T>::Node(){
    m_tpData = new T;
    m_tpNext = NULL;
}
//　帶參數(shù)的構(gòu)造函數(shù)
template<typename T>
Node<T>::Node<T data>{
    m_tpData = new T(data);
    m_tpNext = NULL;
}

template<typename T>
Node<T>::~Node(){
    delete m_tpData;
    m_tpNext = NULL;
}

template<typename T>
void Node<T>::setData(T data){
    *m_tpData = data;
}

template<typename T>
T Node<T>::getData(){
    return *m_tpData;
}

template<typename T>
void Node<T>::setNext(Node<T>* next){
    m_tpNext = next;
}

template <typename T>
Node<T>* Node<T>::getNext(){
    return m_tpNext;
}

template <typename T>
void Node<T>::printData(){
    cout << *m_tpData << endl;
}

template<typename T>
class LinkList{
public:
    LinkList();
    ~LinkList();
    bool isListEmpty();
    bool clearList();
    int getListLength();
    int getElemIndex(T &elem);
    bool getListElem(int index,T* elem);

    bool ListInsert(int index,T &elem);
    bool ListDelete(int index,T *elem);
    void ListPrint(void);
private:
    Node<T>* m_pList;
    int m_iLength;
}

template<typename T>
LinkList<T>::LinkList(){
    m_pList = new Node<T>;
    m_pList->setData(NULL);
    m_pList->setNext(NULL);
    m_iLength = 0;
}

template <typename T>
LinkList<T>::~LinkList(){
    Node<T> *nextNode = m_pList;
    while( nextNode -> getNext() != NULL){
        nextNode = m_pList -> getNext();
        delete m_pList;
        m_pList = nextNode;
    }
    delete m_pList;
    m_pList = NULL;
}

template <typename T>
bool LinkList<T>::isListEmpty(){
    return m_iLength == 0 ? true : false;
}

template <typename T>
bool LinkList<T>::clearList(){
    if(isListEmpty()){
        cout << "List is empty, clear fail." << endl;
    }
    // delete all nodes except the first one
    Node<T>* nowNode = m_pList->getNext();
    Node<T>* nextNode = m_pList->getNext();
    while(nextNode -> getNext() != NULL){
        nextNode = nowNode -> getNext();
        delete nowNode;
        nowNode = nextNode;
    }
    delete nowNode;

    // reset the list length
    m_iLength = 0;
    m_pList -> setNext(NULL);
    return true;
}

template <typename T>
int LinkList<T>::getListLength()
{
    return m_iLength;
}

template <typename T>
int LinkList<T>::getElemIndex(T &elem){
    //　獲取元素的索引
    Node<T>* tempNode = m_pList;
    for(int i=0; i<m_iLength; i++){
        tempNode = tempNode -> getNext();
        if( tempNode-> getData() == elem ) return i;
    }
    // 沒有找到
    return -1;
}

template <typename T>
bool LinkList<T>::getListElem(int index,T* elem)｛
        if( index < 0 || index >= m_iLength){
            return false;
        }

        Node<T>* tempNode = m_pList;
        for( int i=0; i<index; i++){
            tempNode = tempNode -> getNext();
        }
        *ele = tempNode -> getData();
        return true;
｝

Leetcode上也有許多與鏈表相關的問題，我們來一一擊破

Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head.

Example:
Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

這道題目是典型的雙指針問題居触，我們只要設置兩個指針就很容易完成。

eg:
array = {1, 2, 3, 4, 5, 6} n = 2 
initial phase
array           1 2 3 4 5 6
pre             ^       
tempNode        ^

offset tempNode  n steps
array           1 2 3 4 5 6
pre             ^       
tempNode            ^

both pointers start to move forward until tempNode meets the end
array           1 2 3 4 5 6
pre                   ^       
tempNode                  ^

do pre->next = pre->next->next to delete the element 

array           1 2 3 4 6
pre                     ^       
tempNode                ^

代碼如下, 要注意一下邊界條件的判斷

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (!head->next) return NULL;
        // baecause n is always valid, so ignore the judge
        ListNode* tempNode = head, *pre = head;
        // offset tempNode 
        for(int i=0; i<n; i++){
            tempNode = tempNode -> next;
        }
        // 如果已經(jīng)移到NULL上
        if ( !tempNode ) return head->next;
        // go through the list together
        while( tempNode  != NULL ){
            if( tempNode -> next == NULL ){
                pre -> next= pre -> next -> next;
            }
                tempNode = tempNode -> next;
                pre = pre -> next;
          
        }
    return head;
    }
};

Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6

暴力的解法就是直接每次都對k個鏈表進行遍歷并且比較老赤，拿到當前鏈表頭上最小的數(shù)進行插入轮洋，這種方法時間復雜度應該是O(n^2)，按照leetcode的尿性肯定會超時抬旺，所以就直接拋棄掉弊予。
下面可以想想一些典型的O(nlgn)的思路。