leetcode:
11
class Solution(object):
def maxArea(self, height):
"""
:type height: List[int]
:rtype: int
"""
nums = height
left = 0
right = len(nums) -1
area = []
while left < right:
area.append(min([nums[left],nums[right]]) * (right-left))
if nums[left] <= nums[right]:
left += 1
else:
right -= 1
return max(area)
思路：用左右两个指针往中间夹逼
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class Solution(object):
def intToRoman(self, num):
out = []
def getN (num,str1):
if num == 0:
out.append(str1)
return
if num >= 1000:
getN(num-1000,str1+'M')
elif num >= 900:
getN(num-900,str1+"CM")
elif num >= 500:
getN(num-500,str1+"D")
elif num >=400:
getN(num-400,str1+"CD")
elif num>=100:
getN(num-100,str1+'C')
elif num>=90:
getN(num-90,str1+"XC")
elif num >= 50:
getN(num-50,str1+"L")
elif num>=40:
getN(num-40,str1+"XL")
elif num >= 10:
getN(num-10,str1+"X")
elif num == 9:
out.append(str1+"IX")
return
elif num >=5:
getN(num-5,str1+"V")
elif num == 4:
out.append(str1+"IV")
return
else:
getN(num-1,str1+"I")
getN(num,'')
return out[0]
采用递归配合if判断
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class Solution(object):
def isValidSudoku(self, board):
seti = []
# 检查行
flag1 = 0
for i in board:
for j in i:
if j != '.':
seti.append(int(j))
i = list(set(seti))
if len(i) != len(seti):
print(i)
flag1 = 1
seti = []
# 检查列：
rotated = list(zip(*board[::-1]))# 翻转
flag2 = 0
seti = []
for i in rotated:
for j in i:
if j != '.':
seti.append(int(j))
i = list(set(seti))
if len(i) != len(seti):
print(i)
flag2 = 1
seti = []
# 检查块：
flag3 = 0
# 分块
start = [[0,0],[0,3],[0,6],[3,0],[3,3],[3,6],[6,0],[6,3],[6,6]]
fenkuai = []
for item in start:
kuai = []
x = item[0]
y = item[1]
endx = x+2
endy = y+2
while x <= endx:
while y <= endy:
kuai.append(board[x][y])
y+=1
y = item[1]
x+=1
fenkuai.append(kuai)
print(fenkuai)
seti = []
for i in fenkuai:
for j in i:
if j != '.':
seti.append(int(j))
i = list(set(seti))
if len(i) != len(seti):
print(i)
flag3 = 1
seti = []
if flag1==0 and flag2==0 and flag3==0:
return True
else:
return False
1、先判断每行是否有重复的，利用的是set去重后比较长度来判断，
2、然后翻转二维list，重复上述过程 1
3、把各个分块当作一行，9个分块就是9行。对分好块后的二维list执行过程 1
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class Solution(object):
def combinationSum(self, candidates, target):
nums = candidates
out = []
def choose(tar,i,sum,tmp):
if sum > tar or i == len(nums):
return
if sum == tar:
out.append(tmp)
return
choose(tar,i,sum+nums[i],tmp+[nums[i]])
choose(tar,i+1,sum,tmp)
choose(target,0,0,[])
return out
利用递归来解决
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class Solution(object):
def plusOne(self, digits):
mystr = ''
for i in digits:
mystr += str(i)
mynum = int(mystr)
mynum += 1
mystr2 = str(mynum)
list2 = list(mystr2)
for i in range(len(list2)):
list2[i] = int(list2[i])
digits =list2
return digits
1、转成字符相加成字符串
2、转成int加一
3、再转成str的list
4、把str的list转成int 的list
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class Solution(object):
def hammingWeight(self, n):
flag = 0
while n!= 0:
if n%2 == 1:
flag += 1
n = n/2
return flag
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350题目描述
class Solution(object):
def intersect(self, nums1, nums2):
mydict = {}
out = []
if len(nums1) <= len(nums2):
miniNum = nums1
maxNum = nums2
else:
miniNum = nums2
maxNum = nums1
for item in miniNum:
if mydict.get(item,'no') == 'no': #没有
mydict[item] = 1
else:
mydict[item] += 1
for item in maxNum:
if mydict.get(item,'no') == 'no':
pass
else:
if mydict[item] != 0:
out.append(item)
mydict[item] -= 1
return out
先找出两个list中那个比较短，然后把短的这个记录进dict中，再循环长的list，看值是否在dict中。
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