鏈表
鏈表和數(shù)組最大的區(qū)別在于，鏈表不支持隨機(jī)訪問(wèn)晨雳，不像數(shù)組可以對(duì)任意一位的數(shù)據(jù)進(jìn)行訪問(wèn)，鏈表只能從頭一個(gè)一個(gè)往下訪問(wèn)奸腺，尋找下一個(gè)元素餐禁，像穿針引線似的。也正因?yàn)殒湵淼倪@種特點(diǎn)突照，增大了鏈表題目的難度帮非。
本文主要討論的是單鏈表，單鏈表中的鏈表結(jié)點(diǎn)結(jié)構(gòu)體定義如下
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
由上面的代碼可以看出绷旗，每個(gè)結(jié)點(diǎn)包含兩個(gè)域喜鼓，一個(gè)是val表示結(jié)點(diǎn)的值副砍，一個(gè)是next指針，指向下一個(gè)結(jié)點(diǎn)庄岖。
1）做鏈表類(lèi)題目的關(guān)鍵是畫(huà)圖豁翎，將鏈表指針指向的內(nèi)存的變化用圖表示出來(lái)。
2）鏈表類(lèi)題目還有一個(gè)重要的技巧是設(shè)置虛擬頭結(jié)點(diǎn)隅忿。在做題的時(shí)候心剥，頭結(jié)點(diǎn)(head)常常要特殊處理，因?yàn)閔ead結(jié)點(diǎn)沒(méi)有前一個(gè)結(jié)點(diǎn)背桐，這就決定了頭節(jié)點(diǎn)和后續(xù)節(jié)點(diǎn)的處理方法不同优烧。但是如果在頭結(jié)點(diǎn)前加了一個(gè)虛擬頭節(jié)點(diǎn)，那它就可以用正常的方法來(lái)處理了链峭。
虛擬頭節(jié)點(diǎn)定義如下：
ListNode* dummyHead = new ListNode(0); //新建一個(gè)結(jié)點(diǎn)畦娄，value為0
dummyHead->next = head; //讓它next指針指向頭結(jié)點(diǎn)
相關(guān)題目
- leetcode #21. Merge Two Sorted Lists 合并兩個(gè)鏈表
- leetcode #206. Reverse Linked List 翻轉(zhuǎn)鏈表
- leetcode #92. Reverse Linked List II 翻轉(zhuǎn)鏈表中指定區(qū)間的結(jié)點(diǎn)
- leetcode #83. Remove Duplicates from Sorted List 刪除鏈表中重復(fù)元素
- leetcode #2. Add Two Numbers 將兩個(gè)鏈表轉(zhuǎn)化成整數(shù)并相加，再存入鏈表中
- leetcode #445. Add Two Numbers II** 和上題類(lèi)似弊仪，但更難一些
- leetcode #237. Delete Node in a Linked ListDelete Node in a Linked List 刪除指定結(jié)點(diǎn)
- leetcode #328. Odd Even Linked List 奇偶鏈表
- leetcode #19. Remove Nth Node From End of List 刪除倒數(shù)第n個(gè)結(jié)點(diǎn)
leetcode #21. Merge Two Sorted Lists 合并兩個(gè)鏈表
題目：Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
if(l1->val > l2->val){
ListNode *tmp = l2;
tmp->next = mergeTwoLists(l1, l2->next);
return tmp;
}
else{
ListNode *tmp = l1;
tmp->next = mergeTwoLists(l1->next, l2);
return tmp;
}
}
};
leetcode #206. Reverse Linked List 反轉(zhuǎn)列表
思路：一般鏈表的題目都會(huì)約束不能對(duì)鏈表的值進(jìn)行操作熙卡，只能對(duì)鏈表的結(jié)點(diǎn)進(jìn)行操作。處理鏈表問(wèn)題励饵，需要多用幾個(gè)指針來(lái)存儲(chǔ)結(jié)點(diǎn)信息驳癌。
在鏈表中，一旦訪問(wèn)鏈表中的某一個(gè)域役听，一定要保證這個(gè)域不為空颓鲜，為了避免這個(gè)問(wèn)題，可以在剛開(kāi)始的判斷一下該鏈表是否為空典予。
Solution:
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* cur = head;
ListNode* pre = NULL;
while(cur!= NULL){
ListNode* next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
return pre;
}
};
leetcode #92. Reverse Linked List II 翻轉(zhuǎn)鏈表中指定區(qū)間的結(jié)點(diǎn)
一定要畫(huà)圖甜滨！題目要求one-pass，所以遍歷的時(shí)候要保存四個(gè)結(jié)點(diǎn)：m前一個(gè)node, 第m個(gè)node, 第n個(gè)node,第n+1個(gè)node;
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
if (head == NULL or head->next == NULL)
return head;
ListNode *dummyHead = new ListNode(0);
dummyHead->next = head;
ListNode *pre = dummyHead;
ListNode *cur = head;
int pos = 1;
while (pos < m) {
pre = cur;
cur = cur->next;
++pos;
}
ListNode *preNode = pre; //m-1 node
ListNode *curNode = cur; //m node
// cur -> Node_m
while (pos <= n) {
ListNode *next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
++pos;
}
preNode->next = pre; // m - 1 -> n
curNode->next = cur; // m -> n + 1
return dummyHead->next;
}
};
leetcode #83. Remove Duplicates from Sorted List 刪除鏈表中重復(fù)元素
無(wú)非就是在遍歷的時(shí)候熙参，用三個(gè)指針來(lái)保存結(jié)點(diǎn)：previous, current, next;
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(head == NULL || head->next == NULL) return head;
ListNode* pre = head;
ListNode* cur = pre->next;
while(cur != NULL)
{
if(cur->val == pre->val)
{
if(cur->next == NULL)
{
pre->next = cur->next;
return head;
}
else
{
cur = cur->next;
pre->next = cur;
}
}
else
{
pre = cur;
cur = cur->next;
}
}
return head;
}
};
leetcode #2. Add Two Numbers 相加艳吠，再存入鏈表中
這題不能用一個(gè)整數(shù)來(lái)保存然后再相加麦备，因?yàn)闀?huì)出現(xiàn)溢出的情況孽椰；下面的答案是逐位相加，將進(jìn)位保存
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *head = new ListNode(0);
ListNode *cur = head;
int extra = 0;
while(l1 || l2 || extra)
{
int sum = (l1?l1->val:0) + (l2?l2->val:0) + extra;
extra = sum/10;
l1 = l1?l1->next:l1;
l2 = l2?l2->next:l2;
cur->next = new ListNode(sum%10);
cur = cur->next;
cout<<sum<<endl;
}
return head->next;
}
};
leetcode #445. Add Two Numbers II
和#2題類(lèi)似凛篙，但是難了很多黍匾。如果把鏈表翻轉(zhuǎn)之后，就轉(zhuǎn)換成了2呛梆，把加完之后的鏈表再翻轉(zhuǎn)就是想要的答案了锐涯。所以該題是83和2的結(jié)合。
class Solution {
public:
ListNode* reverse(ListNode* head)
{
ListNode* pre = NULL;
ListNode* cur = head;
if(cur == NULL) return head;
while(cur != NULL) {
ListNode* next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
return pre;
}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
l1 = reverse(l1);
l2 = reverse(l2);
ListNode head(0);
ListNode* cur = &head;
int extra = 0;
while(l1 || l2 || extra){
int sum = (l1?l1->val:0 )+(l2?l2->val:0) + extra;
extra = sum/10;
l1 = l1?l1->next:0;
l2 = l2?l2->next:0;
cur->next = new ListNode(sum%10);
cur = cur->next;
}
cur = reverse(head.next);
return cur;
}
};
leetcode #237. Delete Node in a Linked ListDelete Node in a Linked List 刪除指定結(jié)點(diǎn)
注意本題是刪除指定節(jié)點(diǎn)填物，而不是指定值
思路：由于指定節(jié)點(diǎn)前一個(gè)節(jié)點(diǎn)無(wú)法得到纹腌，所以就將下一個(gè)元素的值覆蓋該節(jié)點(diǎn)霎终，然后將下一個(gè)節(jié)點(diǎn)刪除。一般情況下升薯，我們對(duì)鏈表操作時(shí)莱褒，都不改變節(jié)點(diǎn)的值，只對(duì)節(jié)點(diǎn)本身操作涎劈。但是這題是特殊情況广凸，所以鏈表問(wèn)題不一定都是穿針引線的題
易錯(cuò)點(diǎn)：要考慮刪除節(jié)點(diǎn)的下一個(gè)節(jié)點(diǎn)為空的情況，這種情況直接刪掉該元素就可以了蛛枚。
class Solution {
public:
void deleteNode(ListNode* node) {
if(node == NULL) return;
if(node->next == NULL) {
delete node;
return;
}
node->val = node->next->val;
ListNode* delNode = node->next;
node->next = delNode->next;
delete delNode;
return;
}
};
leetcode #328. Odd Even Linked List 奇偶鏈表
class Solution {
public:
ListNode* oddEvenList(ListNode* head) {
if(head == NULL || head->next == NULL) return head;
if(head->next->next == NULL) return head;
ListNode* preOdd = head;
ListNode* preEven = head->next;
ListNode* odd = head->next->next;
//目前肯定有>=3個(gè)節(jié)點(diǎn)
ListNode* even = head->next->next->next;
preOdd->next = odd;
odd->next = preEven;
preEven->next = even;
if(preEven->next == NULL){
return preOdd;
}
//目前肯定有4個(gè)節(jié)點(diǎn);
int i = 1;
ListNode* cur = even->next; //第5個(gè)節(jié)點(diǎn)
while(cur != NULL)
{
if(i%2 != 0)
{
odd->next = cur;
cur = cur->next;
odd = odd->next;
odd->next = preEven;
}
else
{
even->next = cur;
cur = cur->next;
even = even->next;
even->next = NULL;
}
i++;
}
even->next = NULL;
return preOdd;
}
};
leetcode #19. Remove Nth Node From End of List 刪除倒數(shù)第n個(gè)結(jié)點(diǎn)
刪除指定倒數(shù)第n個(gè)元素
思路：解法1）就是two-pass;第一遍遍歷鏈表長(zhǎng)度谅海，第二遍刪除元素
解法2）保存p節(jié)點(diǎn)和q節(jié)點(diǎn)，p和q之間對(duì)節(jié)點(diǎn)數(shù)剛好為n蹦浦，相當(dāng)于一個(gè)滑動(dòng)窗口扭吁，不斷往后移動(dòng)，直到q節(jié)點(diǎn)移動(dòng)到空節(jié)點(diǎn)時(shí)盲镶，這時(shí)p節(jié)點(diǎn)指向的節(jié)點(diǎn)時(shí)倒數(shù)n個(gè)節(jié)點(diǎn)的前一個(gè)節(jié)點(diǎn)
易錯(cuò)點(diǎn)：弄清題目中的n是從0開(kāi)始還是從1開(kāi)始算的智末；在這題中n是從1開(kāi)始數(shù)，并且n是合法的
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummyHead = new ListNode(0);
dummyHead->next = head;
ListNode* p = dummyHead;
ListNode* cur = head;
while(n--)
{
cur = cur->next;
}
ListNode* q = cur;
while(q != NULL)
{
p = p->next;
q = q->next;
}
ListNode* delNode = p->next;
p->next = delNode->next;
delete delNode;
return dummyHead->next;
}
};
