連結串列題目(java實現)
阿新 • • 發佈：2019-01-06
廢話少說，上鍊表的資料結構
class ListNode {
ListNode next;
int val;
ListNode(int x){
val = x;
next = null;
}
}
1.翻轉連結串列
ListNode reverse(ListNode node){ ListNode prev = null; while(node!=null){ ListNode tmp = node.next; node.next = prev; prev = node; node = tmp; } return prev; } //翻轉連結串列(遞迴方式) ListNode reverse2(ListNode head){ if(head.next == null){ return head; } ListNode reverseNode = reverse2(head.next); head.next.next = head; head.next = null; return reverseNode; }
2.判斷連結串列是否有環
boolean hasCycle(ListNode head){ if(head == null|| head.next == null){ return false; } ListNode slow,fast; fast = head.next; slow = head; while(fast!=slow){ if(fast==null||fast.next==null){ return false; } fast = fast.next.next; slow = slow.next; } return true; }
3,連結串列排序
ListNode sortList(ListNode head){ if(head == null|| head.next == null){ return head; } ListNode mid = middleNode(head); ListNode right = sortList(mid.next); mid.next = null; ListNode left = sortList(head); return merge(left, right); } ListNode middleNode(ListNode head){ ListNode slow = head; ListNode fast = head.next; while(fast!=null&fast.next!=null){ slow = slow.next; fast = fast.next.next; } return slow; } ListNode merge(ListNode n1,ListNode n2){ ListNode dummy = new ListNode(0); ListNode node = dummy; while (n1!=null&&n2!=null) { if(n1.val<n2.val){ node.next = n1; n1 = n1.next; }else{ node.next = n2; n2 = n2.next; } node = node.next; } if(n1!=null){ node.next = n1; }else{ node.next = n2; } return dummy.next; }
4.連結串列相加求和
ListNode addLists(ListNode l1,ListNode l2){
if(l1==null&&l2==null){
return null;
}
ListNode head = new ListNode();
ListNode point = head;
int carry = 0;
while(l1!=null&&l2!=null){
int sum = carry + l1.val + l2.val;
point.next = new ListNode(sum%10);
carry = sum/10;
l1 = l1.next;
l2 = l2.next;
point = point.next;
}
while(l1!=null){
int sum = carry + l1.val;
point.next = new ListNode(sum%10);
carry = sum/10;
l1 = l1.next;
point = point.next;
}
while(l2!=null){
int sum = carry + l2.val;
point.next = new ListNode(sum%10);
carry = sum/10;
l2 = l2.next;
point = point.next;
}
if(carry!=0){
point.next = new ListNode(carry);
}
return head.next;
}
5.得到連結串列倒數第n個節點
ListNode nthToLast(ListNode head,int n ){
if(head == null||n<1){
return null;
}
ListNode l1 = head;
ListNode l2 = head;
for(int i = 0;i<n-1;i++){
if(l2 == null){
return null;
}
l2 = l2.next;
}
while(l2.next!=null){
l1 = l1.next;
l2 = l2.next;
}
return l1;
}
6.刪除連結串列倒數第n個節點
ListNode deletenthNode(ListNode head,int n){
// write your code here
if (n <= 0) {
return null;
}
ListNode dumy = new ListNode(0);
dumy.next = head;
ListNode prdDel = dumy;
for(int i = 0;i<n;i++){
if(head==null){
return null;
}
head = head.next;
}
while(head!=null){
head = head.next;
prdDel = prdDel.next;
}
prdDel.next = prdDel.next.next;
return dumy.next;
}
7.刪除連結串列中重複的元素
ListNode deleteMuNode(ListNode head){
if(head == null){
return null;
}
ListNode node = head;
while(node.next != null){
if(node.val == node.next.val){
node.next = node.next.next;
}else{
node = node.next;
}
}
return head;
}
8.刪除連結串列中重複的元素ii,去掉重複的節點
ListNode deleteMuNode2(ListNode head){
if(head == null||head.next == null){
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
while(head.next!=null&&head.next.next!=null){
if(head.next.val == head.next.next.val){
int val = head.next.val;
while(head.next.val == val&&head.next != null){
head.next = head.next.next;
}
}else{
head = head.next;
}
}
return dummy.next;
}
9.旋轉連結串列
ListNode rotateRight(ListNode head,int k){
if(head ==null){
return null;
}
int length = getLength(head);
k = k % length;
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
ListNode tail = dummy;
for(int i = 0;i<k;i++){
head = head.next;
}
while(head.next!= null){
head = head.next;
tail = tail.next;
}
head.next = dummy.next;
dummy.next = tail.next;
tail.next = null;
return dummy.next;
}
10.重排連結串列
ListNode reOrder(ListNode head){
if(head == null||head.next == null){
return;
}
ListNode mid = middleNode(head);
ListNode tail = reverse(mid.next);
mergeIndex(head, tail);
}
private void mergeIndex(ListNode head1,ListNode head2){
int index = 0;
ListNode dummy = new ListNode(0);
while (head1!=null&&head2!=null) {
if(index%2==0){
dummy.next = head1;
head1 = head1.next;
}else{
dummy.next = head2;
head2 = head2.next;
}
dummy = dummy.next;
index ++;
}
if(head1!=null){
dummy.next = head1;
}else{
dummy.next = head2;
}
}
11.連結串列劃分
ListNode partition(ListNode head,int x){
if(head == null){
return null;
}
ListNode left = new ListNode(0);
ListNode right = new ListNode(0);
ListNode leftDummy = left;
ListNode rightDummy = right;
while(head!=null){
if(head.val<x){
left.next = head;
left = head;
}else{
right.next = head;
right = head;
}
head = head.next;
}
left.next = rightDummy.next;
right.next = null;
return leftDummy.next;
}
12.翻轉連結串列的n到m之間的節點
ListNode reverseN2M(ListNode head,int m,int n){
if(m>=n||head == null){
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
for(int i = 1;i<m;i++){
if(head == null){
return null;
}
head = head.next;
}
ListNode pmNode = head;
ListNode mNode = head.next;
ListNode nNode = mNode;
ListNode pnNode = mNode.next;
for(int i = m;i<n;i++){
if(pnNode == null){
return null;
}
ListNode tmp = pnNode.next;
pnNode.next = nNode;
nNode = pnNode;
pnNode = tmp;
}
pmNode.next = nNode;
mNode.next = pnNode;
return dummy.next;
}
13.合併K個排序過的連結串列
ListNode mergeKListNode(ArrayList<ListNode> k){
if(k.size()==0){
return null;
}
return mergeHelper(k,0,k.size()-1);
}
ListNode mergeHelper(List<ListNode> lists,int start,int end){
if(start == end){
return lists.get(start);
}
int mid = start + ( end - start )/2;
ListNode left = mergeHelper(lists, start, mid);
ListNode right = mergeHelper(lists, mid+1, end);
return mergeTwoLists(left,right);
}
ListNode mergeTwoLists(ListNode list1,ListNode list2){
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while(list1!=null&&list2!=null){
if(list1.val<list2.val){
tail.next = list1;
tail = tail.next;
list1 = list1.next;
}else{
tail.next = list2;
tail = list2;
list2 = list2.next;
}
}
if(list1!=null){
tail.next = list1;
}else{
tail.next = list2;
}
return dummy.next;
}