點我跳轉題目鏈結
class
solution
else
cur = cur.next;}if
(head.val == val)
return head;
}}
方法一：
//改變前後節點指向
public listnode reverselist
(listnode head)
cur.next = pre;
pre = cur;
cur = curnext;
}return head;
}
//頭插法
public listnode reverselist1 (listnode head)
listnode cur = head.next;
head.next = null;
while
(cur != null)
return head;
}
public listnode middlenode
(listnode head)
return slow;
}
方法一：
//第一次
public listnode findkthtotail
(listnode head,
int k)
while
(fast != null)
if(count < k)
return null;
return slow;
}
//優化
public listnode findkthtotail1
(listnode head,
int k)
else
}while
(fast != null)
return slow;
}
public listnode mergetwolists
(listnode l1, listnode l2)
else
tmp = tmp.next;}if
(l2 == null)
else
head = head.next;
return head;
}
點我跳轉題目鏈結
public listnode partition
(listnode phead,
int x)
else
cur = cur.next;
} a.next = headb.next;
b.next = null;
heada = heada.next;
return heada;
}
public listnode deleteduplication1
(listnode phead)
}else
cur = cur.next;
} newhead = newhead.next;
tmp.next = null;
return newhead;
}
public
boolean
chkpalindrome
(listnode a)
//然後翻轉中間節點後面節點的指向
listnode cur = slow.next;
slow.next =null;
while
(cur != null)
//最後從後往前 以及 從前往後 比較每個節點的值
while
(slow != null)
slow = slow.next;
a = a.next;
}return
true
;}
public listnode getintersectionnode
(listnode heada, listnode headb)
}else
}while
(cura != curb)
return cura;
}public
intsize
(listnode head)
return count;
}
public
boolean
hascycle
(listnode head)
while
(slow != fast)
;return
true
;}
public listnode detectcycle
(listnode head)
while
(slow != fast)
; slow = head;
while
(slow != fast)
return fast;
}
public
void
deletenode
(listnode node)
方法一：
//時間會慢
public listnode rotateright2
(listnode head,
int k)
else
k--;}
if(fast == slow)
return head;
while
(fast.next != null)
listnode tmp = slow.next;
slow.next = null;
fast.next = head;
head = tmp;
return head;
}
//快
public listnode rotateright
(listnode head,
int k)
cur.next = head;
int step = count-k%count;
while
(step !=0)
cur = head;
while
(count !=1)
cur.next = null;
return head;
}
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