看了有助于你面试的单链表的OJ面试题
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目录
1.反转链表
反转链表我们先从前面开始反转，一个一个节点来
public ListNode reverseList()
if(this.head == null)
return null;
ListNode cur = this.head;
ListNode prev = null ;
ListNode curNext = cur.next;
while (cur != null)
cur.next = prev;
prev = cur;
cur = curNext;
curNext = cur.next;
return prev;
2.链表的中间节点
定义一个快慢指针来走，快的指针走的路程比慢的多一倍，所以慢指针指向的就是中间节点。
如果链表节点数是双数则返回第二个中间节点，如下所示
public ListNode middleNode()
if(this.head == null)
return null;
ListNode fast = this.head;
ListNode slow =this.head;
while(fast != null && fast.next != null)
fast = fast.next.next;
slow = slow.next;
return slow;
3.找到倒数第 k 个节点
public ListNode FindKthToTail(int k)
if(k <= 0 || head ==null)
return null;
ListNode fast = this.head;
ListNode slow = this.head;
while(k-1 != 0)
fast = fast.next;
//fast == null是防止求的倒数第k个数超出节点长度，
if(fast == null)
return null;
k--;
while(fast.next != null)
fast = fast.next;
slow = slow.next;
return slow;
4.分割链表
分割链表。给你一个链表的头节点 head 和一个特定值 x ， 使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
public ListNode partition(int x)
ListNode bs = null;
ListNode be = null;
ListNode as = null;
ListNode ae = null;
ListNode cur = head;
while (cur != null)
if(cur.val < x)
//第一次
if(bs == null)
bs = cur;
be = cur;
else
//不是第一次
be.next = cur;
be = be.next;
else
//第一次
if(as == null)
as = cur;
ae = cur;
else
ae.next = cur;
ae = ae.next;
cur = cur.next;
//预防第一段为空
if(bs == null)
return as;
be.next = as;
//预防最后一个节点的next域不为空
if(as != null)
ae.next = null;
return bs;
5.删除链表中的重复节点
升序排列的链表，删除多余重复的元素，使每个元素 只出现一次
public ListNode deleteDuplication()
ListNode cur = head;
ListNode newHead = new ListNode(-1);
ListNode tmp = newHead;
while (cur != null)
if(cur.next != null && cur.val == cur.next.val)
//写while循环是防止有两个以上相同的
while (cur.next != null && cur.val == cur.next.val)
cur = cur.next;
// cur = cur.next;//这一步是删完相同的节点，不留下一个
else
tmp.next = cur;
tmp = tmp.next;
cur = cur.next;
tmp.next = null;
return newHead.next;
6.判断回文链表
public boolean isPalindrome()
ListNode prev = head;
ListNode fast = head;
ListNode slow = head;
//1.slow找到了中间位置
while (fast != null && fast.next != null)
fast = fast.next.next;
slow = slow.next;
//2.反转后半部分
ListNode cur = slow.next;
while (cur != null)
ListNode curNext = cur.next;
cur.next = slow;
slow = cur;
cur = curNext;
//3.两边往中间走，判断回文
while (prev != slow)
if(slow.val != prev.val)
return false;
if(prev.next == slow)
return true;
slow = slow.next;
prev = prev.next;
return true;
7.判断链表是否有环
public boolean hasCycle()
if(head == null) return false;
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null)
fast = fast.next.next;
slow = slow.next;
if (fast == slow)
return true;
return false;
8.一个环形链表，返回链表开始入环的第一个节点
public ListNode detectCycle(ListNode head)
if(head == null) return null;
ListNode fast = head;
ListNode slow = head;
//判断是否有环
while (fast != null && fast.next != null)
fast = fast.next.next;
slow = slow.next;
if (fast == slow)
break;//相遇退出循环
if (fast == null || fast.next == null)
return null;
//头和相遇点到环的入口点距离相等
fast = head;
while (fast != slow)
fast = fast.next;
slow = slow.next;
return fast;
9.找两个链表相交的节点
两个链表相交，及next 域相同
public ListNode getIntersectionNode(ListNode headA, ListNode headB)
if(headA == null || headB == null)
return null;
ListNode pl = headA;
ListNode ps = headB;
int lenA = 0;
int lenB = 0;
while (pl != null)
lenA++;
pl = pl.next;
pl = headA;
while (ps != null)
lenB++;
ps = ps.next;
ps = headB;
int len = lenA - lenB;
if(len < 0)
pl = headB;
ps = headA;
len = lenB - lenA;
//pl走差值len步
while (len != 0)
pl = pl.next;
len--;
//同时走，直到相遇
while (pl != ps)
pl = pl.next;
ps = ps.next;
//返回相遇的节点
return pl;//有值就返回值，是空就返回空。
10.合并两个顺序链表
public static ListNode mergeTwoLists(ListNode headA,ListNode headB)
ListNode newHead = new ListNode(-1);
ListNode tmp = newHead;
while(headA != null && headB != null)
if(headA.val < headB.val)
tmp.next = headA;
headA = headA.next;
tmp = tmp.next;
else
tmp.next = headB;
headB = headB.next;
tmp = tmp.next;
if(headA != null)
tmp.next = headA;
if(headB != null)
tmp.next = headB;
return newHead.next;
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