Leetcode 演算法題07

169. majority element

輸入乙個列表，找出其中出現次數超過列表長度一半的數

我的**：

class solution(object):

def majorityelement(self, nums):
""":type nums: list[int]
:rtype: int
"""count = collections.counter(nums)
for i in count:
if count[i] > len(nums)/2:
return i

class solution(object):

def majorityelement(self, nums):
""":type nums: list[int]
:rtype: int
"""return sorted(nums)[len(nums)/2]

167. two sum ii - input array is sorted

給乙個排好序的列表和目標數，找出列表中相加得到目標數的兩個數的索引

input:

numbers=, target=9

output:

index1=1, index2=2

我的**：第一次超時了，沒有記錄，這次是想了個不超時的思路，寫得多了點

class solution(object):

def twosum(self, numbers, target):
""":type numbers: list[int]
:type target: int
:rtype: list[int]
"""number = collections.counter(numbers)
print(number)
keys = list(number.keys())
keys.sort()
print(keys)
for i in range(len(keys)):
if keys[i] > target:
return false
add = 0
while i+add 1:
ans1 = sum(list(map(lambda x:number[x],keys[:i])))+1
ans2 = ans1+1
return [ans1,ans2]

# two-pointer

def twosum1(self, numbers, target):
l, r = 0, len(numbers)-1
while l < r:
s = numbers[l] + numbers[r]
if s == target:
return [l+1, r+1]
elif s < target:
l += 1
else:
r -= 1
# dictionary           
def twosum2(self, numbers, target):
dic = {}
for i, num in enumerate(numbers):
if target-num in dic:
return [dic[target-num]+1, i+1]
dic[num] = i
# binary search        
def twosum(self, numbers, target):
for i in xrange(len(numbers)):
l, r = i+1, len(numbers)-1
tmp = target - numbers[i]
while l <= r:
mid = l + (r-l)//2
if numbers[mid] == tmp:
return [i+1, mid+1]
elif numbers[mid] < tmp:
l = mid+1
else:
r = mid-1

387. first unique character in a string

輸入乙個字串，找到第乙個在字串中沒有重複出現過的字母的索引

examples:

s = "leetcode"

return 0.
s = "loveleetcode",
return 2.

class solution(object):

def firstuniqchar(self, s):
""":type s: str
:rtype: int
"""a=collections.counter(s)
ans = 
for i in a:
if a[i] == 1:
if ans == :
return -1
return min([s.find(j) for j in ans])

def firstuniqchar(self, s):

""":type s: str
:rtype: int
"""letters='abcdefghijklmnopqrstuvwxyz'
index=[s.index(l) for l in letters if s.count(l) == 1]
return min(index) if len(index) > 0 else -1

237. delete node in a linked list

刪除節點

我的**：感覺有時候出現的問題好蠢

# definition for singly-linked list.

# class listnode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = none
class solution(object):
def deletenode(self, node):
""":type node: listnode
:rtype: void do not return anything, modify node in-place instead.
"""node.val ,node.next= node.next.val,node.next.next

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