Linked List
445 Add Two Numbers II
思路：利用stack；或者递归计算 并用一个进位来计算
328 Odd Even Linked List
public ListNode oddEvenList(ListNode head) {
if(head==null||head.next==null) return head;
ListNode odd=head,ehead=head.next,even=ehead;
while(even!=null&&even.next!=null){
odd.next=even.next;
odd=odd.next;
even.next=odd.next;
even=even.next;
}
odd.next=ehead;
return head;
}
206 Reverse Linked List
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseList(ListNode head) {
ListNode newHead=null;
while(head!=null){
ListNode node=head.next; //主要是为了记录下一个node，保证链表不会断掉
head.next=newHead;
newHead=head;
head=node;
}
return newHead;
}
}
203 Remove Linked List Elements
Remove all elements from a linked list of integers that have value val.
Example
*Given:* 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
*Return:* 1 --> 2 --> 3 --> 4 --> 5
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
本来想利用删除元素的办法来做，貌似结尾有点问题，可以用两个临时变量。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode fakeHead = new ListNode(-1);
fakeHead.next = head;
ListNode curr = head, prev = fakeHead;
while (curr != null) {
if (curr.val == val) {
prev.next = curr.next;
} else {
prev = prev.next;
}
curr = curr.next;
}
return fakeHead.next;
}
}
160 Intersection of Two Linked Lists
很多办法：
先计算长度，从同一位置往后走，直到val相等
148 Sort List
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
//归并，分割连表；分别排序；排序
public ListNode sortList(ListNode head) {
if(head==null ||head.next==null){
return head;
}
ListNode prev = null;
ListNode slow=head;
ListNode fast=head;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
prev.next=null;
// step 2. sort each half
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
// step 3. merge l1 and l2
return merge(l1, l2);
}
ListNode merge(ListNode l1, ListNode l2) {
ListNode l = new ListNode(0), p = l;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
if (l1 != null)
p.next = l1;
if (l2 != null)
p.next = l2;
return l.next;
}
}
147 Insertion Sort List
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode insertionSortList(ListNode head) {
if( head == null ){
return head;
}
ListNode helper = new ListNode(0); //new starter of the sorted list
ListNode cur = head; //the node will be inserted
ListNode pre = helper; //insert node between pre and pre.next
ListNode next = null; //the next node will be inserted
//not the end of input list
while( cur != null ){
next = cur.next;
//find the right place to insert
while( pre.next != null && pre.next.val < cur.val ){
pre = pre.next;
}
//insert between pre and pre.next
cur.next = pre.next;
pre.next = cur;
pre = helper;
cur = next;
}
return helper.next;
}
}
143 Reorder List
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void reorderList(ListNode head) {
if(head==null||head.next==null) return;
//Find the middle of the list
ListNode p1=head;
ListNode p2=head;
while(p2.next!=null&&p2.next.next!=null){
p1=p1.next;
p2=p2.next.next;
}
//Reverse the half after middle 1->2->3->4->5->6 to 1->2->3->6->5->4
ListNode preMiddle=p1;
ListNode preCurrent=p1.next;
while(preCurrent.next!=null){
ListNode current=preCurrent.next;
preCurrent.next=current.next;
current.next=preMiddle.next;
preMiddle.next=current;
}
//Start reorder one by one 1->2->3->6->5->4 to 1->6->2->5->3->4
p1=head;
p2=preMiddle.next;
while(p1!=preMiddle){
preMiddle.next=p2.next;
p2.next=p1.next;
p1.next=p2;
p1=p2.next;
p2=preMiddle.next;
}
}
}
141 Linked List Cycle
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
快慢指针 如果有环 一定会相遇
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* Use two pointers, walker and runner.
walker moves step by step. runner moves two steps at time.
if the Linked List has a cycle walker and runner will meet at some
point.
*/
public boolean hasCycle(ListNode head) {
if(head==null) return false;
ListNode walker = head;
ListNode runner = head;
while(runner.next!=null && runner.next.next!=null) {
walker = walker.next;
runner = runner.next.next;
if(walker==runner) return true;
}
return false;
}
}
142 Linked List Cycle
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
就知道会让求连接点。
假设s步以后，为相交点。 再走m步后相遇， 那么此刻fast比slow夺走了一个环（k个环） 此刻 假如环的长度为x； 2k-k=x; k=s+m;s+m=k; 那么意味着slow指针此刻距离交叉点的距离和链表头节点剧烈交叉点距离相等。分别一次走一步，直到想等。
public ListNode detectCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;
if (fast == slow){
ListNode slow2 = head;
while (slow2 != slow){
slow = slow.next;
slow2 = slow2.next;
}
return slow;
}
}
return null;
}
138 Copy List with Random Pointer
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
思路：
先复制节点（不加radom）； 在 复制radom信息；
最后删除重复的元素，并复制新的连表。
/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/
public class Solution {
public RandomListNode copyRandomList(RandomListNode head) {
RandomListNode iter = head, next;
// First round: make copy of each node,
// and link them together side-by-side in a single list.
while (iter != null) {
next = iter.next;
RandomListNode copy = new RandomListNode(iter.label);
iter.next = copy;
copy.next = next;
iter = next;
}
// Second round: assign random pointers for the copy nodes.
iter = head;
while (iter != null) {
if (iter.random != null) {
iter.next.random = iter.random.next;
}
iter = iter.next.next;
}
// Third round: restore the original list, and extract the copy list.
iter = head;
RandomListNode pseudoHead = new RandomListNode(0);
RandomListNode copy, copyIter = pseudoHead;
while (iter != null) {
next = iter.next.next;
// extract the copy
copy = iter.next;
copyIter.next = copy;
copyIter = copy;
// restore the original list
iter.next = next;
iter = next;
}
return pseudoHead.next;
}
}
109 Convert Sorted List to Binary Search Tree
由于排好序，二叉搜索树的中序遍历是有序的。所以，利用快慢指针，每次返回根节点，递归进行构造。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
if(head==null) return null;
return toBST(head,null);
}
public TreeNode toBST(ListNode head,ListNode tail){
ListNode slow = head;
ListNode fast = head;
if(head==tail) return null;
while(fast!=tail&&fast.next!=tail){
fast = fast.next.next;
slow = slow.next;
}
TreeNode thead = new TreeNode(slow.val);
thead.left = toBST(head,slow);
thead.right = toBST(slow.next,tail);
return thead;
}
}
92 Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ? m ? n ? length of list.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null) return null;
ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
dummy.next = head;
ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
for(int i = 0; i pre = 1, start = 2, then = 3
// dummy-> 1 -> 2 -> 3 -> 4 -> 5
for(int i=0; i1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
// second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
return dummy.next;
}
}