4 鍊錶和遞迴

輸入: 1->2->6->3->4->5->6, val = 6

輸出: 1->2->3->4->5

/**

* definition for singly-linked list.
* public class listnode 
* }*/class
solution
if(head == null)
listnode prev = head;
while(prev.next != null)
else
}return head;
}}

public

class
solution2
else 
}return dummyhead.next;
}}

# definition for singly-linked list.

# class listnode:
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class
solution:
defremoveelements
(self, head, val):
"""        :type head: listnode
:type val: int
:rtype: listnode
"""dummyhead = listnode(-1)
dummyhead.next = head
prev = dummyhead
while prev.next != none:
if prev.next.val == val:
delnode = prev.next
prev.next = delnode.next
delnode.next = none
else:
prev = prev.next
return dummyhead.next

public class listnode 

// 鍊錶節點的建構函式
// 使用arr為引數，
建立乙個鍊錶，
當前的listnode為煉表頭節點
public  listnode(int arr)
this.val = arr[0];
listnode cur = this;
for(int i = 1; i < arr.length; i++)
}// 以當前的節點為頭節點餓的鍊錶資訊字串
@override
public string tostring()
return res.tostring();
}

class

solution
// 測試
public
static
void
main
(string args)
;        listnode head = new listnode(nums);
system.out.println(head);
listnode res = (new solution()).removeelements(head, 6);
system.out.println(res);
}}

1->2->6->3->4->5->6->null

1->2->3->4->5->null

public

class
sum    // 計算arr[l...n）這個區間內所有數字的和
private
static
intsum
(int arr, int l)
return arr[l] + sum(arr, l + 1);  // 把原問題轉化為更小的問題
}// 測試
public
static
void
main
(string args)
;        system.out.println(sum(nums));
}

36

public

class
solution3
head.next = removeelements(head.next, val);
return head.val == val ? head.next : head;
}// 測試
public
static
void
main
(string args)
;        listnode head = new listnode(nums);
system.out.println(head);
listnode res = (new solution3()).removeelements(head, 6);
system.out.println(res);
}}

1->2->6->3->4->5->6->null

1->2->3->4->5->null

public

class
solution3
head.next = removeelements(head.next, val, depth + 1);  // depth表示遞迴深度，每次遞迴都+1
system.out.print(depthstring);
system.out.println("after remove" + val + ":" + head.next);
listnode ret;
if (head.val == val)  else 
system.out.print(depthstring);
system.out.println("return: " + ret);
return ret;
}// 為了列印遞迴深度
private string generatedepthstring
(int depth)
return res.tostring();
}// 測試
public
static
void
main
(string args)
;        listnode head = new listnode(nums);
system.out.println(head);
listnode res = (new solution3()).removeelements(head, 6, 0);
system.out.println(res);
}}

1->2->6->3->4->5->6->null

call: remove6 in
1->2->6->3->4->5->6->null
--call: remove6 in
2->6->3->4->5->6->null
----call: remove6 in
6->3->4->5->6->null
------call: remove6 in
3->4->5->6->null
--------call: remove6 in
4->5->6->null
----------call: remove6 in
5->6->null
------------call: remove6 in
6->null
--------------call: remove6 in null
--------------return: null
------------after remove6: null
------------return: null
----------after remove6: null
----------return: 5->null
--------after remove6: 5->null
--------return: 4->5->null
------after remove6: 4->5->null
------return: 3->4->5->null
----after remove6: 3->4->5->null
----return: 3->4->5->null
--after remove6: 3->4->5->null
--return: 2->3->4->5->null
after remove6: 2->3->4->5->null
return: 1->2->3->4->5->null
1->2->3->4->5->null

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