Happy Number
202. Happy Number
Solution
Use set to remember the visited numbers.
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
visited = set()
while n not in visited:
s = 0
visited.add(n)
while n != 0:
s += (n % 10) ** 2
n = n / 10
if s == 1:
return True
n = s
return False
...
Count and Say
38. Count and Say
Solution
Reference
def countAndSay(self, n):
s = '1'
for _ in range(n - 1):
s = ''.join(str(len(list(group))) + digit
for digit, group in itertools.groupby(s))
return s
Pascal's Triangle
118. Pascal’s Triangle
Solution
def generate(self, numRows):
"""
:type numRows: int
:rtype: List[List[int]]
"""
if numRows == 0:
return []
res = [[1]]
for i in range(2, numRows+1):
row = [1] * i
for j in range(1, i-1):
row[j] = res[-1][j-1] + res[-1][j]
res.append(row)
...
Number of 1 Bits
191. Number of 1 Bits
Solution
def hammingWeight(self, n):
"""
:type n: int
:rtype: int
"""
res = 0
while n != 0:
tmp = n
n = n >> 1
if n << 1 != tmp:
res += 1
return res
Missing Number
268. Missing Number
Solution
Find the real number index in the arr.
Set the number at the arr[index] to a updated number,
which contains both the information that it is set and its original number.
Find the unset number.
def missingNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
N = len(nums)
# pri...
First Unique Character in a String
387. First Unique Character in a String
Solution
Get all indexes of characters in the string.
Can use array and ascii number instead.
If the length of indexes is 1, collect the index.
Find the minimum of them.
def firstUniqChar(self, s):
"""
:type s: str
:rtype: int
"""
dic = collections.defaultdict(lis...
Maximum Nesting Depth of the Parentheses
1614. Maximum Nesting Depth of the Parentheses
Solution
The string is a valid parentheses string. We only need to count( for the number of parentheses.
Remember to update the result if possible.
def maxDepth(self, s):
res = cur = 0
for c in s:
if c == '(':
cur += 1
res = max(res, cur...
Maximal Network Rank
1615. Maximal Network Rank
There is an infrastructure of n cities with some number of roads connecting these cities. Each roads[i] = [ai, bi] indicates that there is a bidirectional road between cities ai and bi.
The network rank of two different cities is defined as the total number of directly connected roads to either city. If a road is dir...
199 post articles, 25 pages.