leetcode 解題報告（21 28）

題目描述：合併兩個有序鍊錶

解題思路：水題

# definition for singly-linked list.
# class listnode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class
solution
(object):
defmergetwolists
(self, l1, l2)
:"""
:type l1: listnode
:type l2: listnode
:rtype: listnode
"""ans =
none
new =
none
while l1 or l2:
ifnot l1:
ifnot new:
new = listnode(0)
ans = new
else
:                    new.
next
= listnode(0)
new = new.
next
new.val = l2.val
l2 = l2.
next
elif
not l2:
ifnot new:
new = listnode(0)
ans = new
else
:                    new.
next
= listnode(0)
new = new.
next
new.val = l1.val
l1 = l1.
next
elif
not l1 and
not l2:
return new
elif l1.val >= l2.val:
ifnot new:
new = listnode(0)
ans = new
else
:                    new.
next
= listnode(0)
new = new.
next
new.val = l2.val
l2 = l2.
next
elif l1.val < l2.val:
ifnot new:
new = listnode(0)
ans = new
else
:                    new.
next
= listnode(0)
new = new.
next
new.val = l1.val
l1 = l1.
next
return ans

題目描述：合併k個有序鍊錶

解題思路：比較難，剛開始用合併兩個有序陣列的方法來做，果斷超時，做了太多的重複計算。後來看到有人說優先佇列，於是只要知道優先佇列就很簡單了，然後去看了一下stl的priority_queue，**就很好寫了，用優先佇列儲存每個鍊錶的當前位置，然後就出隊進隊就行了。

class solution };
listnode*
mergeklists
(vector>
& lists)
else
if(tmp->next)
que.
push
(tmp->next);}
return head;}}
;

題目描述：把鍊錶的每兩個元素調換個位置

解題思路：水題，指標注意好就行了

class solution };
listnode*
mergeklists
(vector>
& lists)
else
if(tmp->next)
que.
push
(tmp->next);}
return head;}}
;

解題思路：難點就在於掌握好每個指標，每組的隊首隊尾都要記錄，還要記錄上一組的隊尾，然後就按照倒置一條鍊錶的思路，用頭插法倒置k個節點，再把當前組和上一組串起來。只要注意好每個指標，思路其實很簡單，花了乙個兩個小時，才順利寫好修改指標的**。

/**
* definition for singly-linked list.
* struct listnode 
* };
*/class solution 
if(cnt == k && end)
else
return head;
while
(end)
fst = end;
mid = end;
int tmp =1;
while
(end && tmp < k)
if(tmp == k && end)
end = mid->next;
else}}
if(grp == head)
head = fst;
else
return head;}}
;

題目描述：移除陣列重複元素

解題思路：水題

class solution 
}return i+1;
}};

題目描述：給定乙個陣列和乙個值k，移除陣列中所有的k

解題思路：水題

class
solution
(object):
defremoveelement
(self, nums, val)
:"""
:type nums: list[int]
:type val: int
:rtype: int
"""length =
len(nums)
-nums.count(val)
while val in nums:
nums.remove(val)
return length

題目描述：定位模式串在正文中的位置

解題思路：kmp入門題

class
solution
(object):
defstrstr
(self, haystack, needle)
:"""
:type haystack: str
:type needle: str
:rtype: int
"""ifnot needle:
return
0        next_ =[0
,0]        p =
0for i in
range(1
,len
(needle)):
p = next_[i]
while p and needle[p]
!= needle[i]
: p = next_[p]
if needle[p]
== needle[i]1)
else0)
# print(next_)
p =0for i in
range
(len
(haystack)):
while p and needle[p]
!= haystack[i]
: p = next_[p]
if needle[p]
== haystack[i]
: p +=
1if p ==
len(needle)
:return i-p+
1return
-1

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