leetcode刷题笔记-math
780. Reaching Points
First while: x, y comes fromx, k*x+y or x+k*y, y for each step
Second: check if we reduce target points to (x, y+kx) or (x+ky, y)
class Solution(object):
def reachingPoints(self, sx, sy, tx, ty):
while sx < tx and sy < ty:
tx, ty = tx % ty, ty % tx
return sx == tx and sy < ty and (ty - sy) % sx == 0 or \
sy == ty and sx < tx and (tx - sx) % sy == 0自己写的TLE的方法：
level = [(sx, sy)]
while True:
Next = []
xmin = float('inf')
ymin = float('inf')
while level:
x, y = level.pop(0)
if (x, y) == (tx, ty):
return True
Next.append((x, x+y))
Next.append((x+y, y))
xmin = min(xmin, x+y)
ymin = min(ymin, y+x)
if xmin > tx and ymin > ty: return False
level = Next
238. Product of Array Except Self
class Solution(object):
def productExceptSelf(self, nums):
# The output array does not count as extra space for the purpose of space complexity analysis.
# 所以res 不算extra space 所以O(1) space
# 先从左到右乘，before的结果是 nums[0]*..*nums[i-1]不包括Nums[i]
# 再从右到左乘，after的结果是nums[i+1]到nums[n-1]的乘积
# 前后相乘就是结果
res = []
before = 1
for i in xrange(len(nums)):
res.append(before)
before *= nums[i]
after = 1
for i in xrange(len(nums)-1, -1, -1):
res[i] *= after
after *= nums[i]
return res
89. Gray Code
class Solution(object):
def grayCode(self, n):
res = [0]
for i in xrange(n):
add = pow(2, i)
for n in reversed(res):
res.append(n+add)
return res
311. Sparse Matrix Multiplication
class Solution(object):
def multiply(self, A, B):
# 无hash table的做法,有的也类似
if not A or not B: return
rowA = len(A)
colA = len(A[0]) # rowB
colB = len(B[0])
re = [[0]*colB for _ in xrange(rowA)]
for i, row in enumerate(A):
for k, eleA in enumerate(row):
if eleA:
for j, eleB in enumerate(B[k]): # K
if eleB:
re[i][j] += eleA * eleB # A[i][k] * B[k][j]
return re296. Best Meeting Point
这题知道要找median的话就是easy level的题目
class Solution(object):
def minTotalDistance(self, grid):
"""
Manhattan Distance, find the median
"""
rowIdx = [i for i in xrange(len(grid)) for j in xrange(len(grid[0])) if grid[i][j]]
colIdx = [j for i in xrange(len(grid)) for j in xrange(len(grid[0])) if grid[i][j]]
colIdx.sort()
row = rowIdx[len(rowIdx) / 2]
col = colIdx[len(rowIdx) / 2]
return sum(abs(i-row)+abs(j-col) for i in xrange(len(grid)) for j in xrange(len(grid[0])) if grid[i][j])227. Basic Calculator II
class Solution(object):
def calculate(self, s):
s += '+'
stack, num, preOp = [], 0, '+'
for c in s:
if c.isdigit():
num = num*10 + int(c)
elif not c.isspace():
if preOp == '-': stack.append(-num)
elif preOp == '+': stack.append(num)
elif preOp == '*':
stack.append(stack.pop() * num)
elif preOp == '/':
n = stack.pop() # stack.append(int(stack.pop() / num)) 不行，因为14-3/2中-3/2 = -2
# *(n/abs(n))如果是负数的话相当于乘以-1 变成正数的除法
stack.append(0 if n == 0 else n * (n/abs(n)) / num * (n/abs(n)))
preOp, num = c, 0
return sum(stack)
224. Basic Calculator
class Solution(object):
def calculate(self, s):
res, num, sign, stack = 0, 0, 1, []
for c in s:
if c.isdigit():
num = num*10 + int(c)
elif c in ['+', '-']:
res += num * sign # 加上前面的num
num = 0
sign = [-1, 1][c == '+']
elif c == '(':
stack.append(res)
stack.append(sign)
res, num, sign = 0, 0, 1 # num其实在 +, -的时候就set为0了
elif c == ')':
res += num*sign
sign = stack.pop()
res = stack.pop() + sign*res
num = 0
return res + num*sign # 最后一个循环里面加不到204. Count Primes
如果2是prime 则x=2*j 的数都不是prime。 2*j < n-1 则 j < (n-1) / 2
class Solution(object):
def countPrimes(self, n):
if n <= 2:
return 0
primes = [True] * n
primes[0] = primes[1] = False
for i in xrange(2, n):
if primes[i] == True:
for j in xrange(2, (n-1)/i + 1):
primes[i*j] = False
return sum(primes)
462. Minimum Moves to Equal Array Elements II
这题只要记住The Median Minimizes the Sum of Absolute Deviations (The L1L1 Norm)
Suppose we have a set SS of real numbers. Show that
∑s∈S|s−x|∑s∈S|s−x|
is minimal if xx is equal to the median
证明：https://math.stackexchange.com/questions/113270/the-median-minimizes-the-sum-of-absolute-deviations-the-l-1-norm
781. Rabbits in Forest
第一次解法，通过，但是写法感觉略复杂，虽然时间复杂度是O（n）
class Solution(object):
def numRabbits(self, answers):
"""
:type answers: List[int]
:rtype: int
"""
if not answers:
return 0
res = 0
answers = sorted(answers)
rest = 0
for i in xrange(len(answers)):
if i == 0:
res += answers[i] + 1
rest = answers[i]
else:
if answers[i] == 0:
res += 1
elif answers[i] != answers[i-1]:
res += answers[i] + 1
rest = answers[i]
elif answers[i] == answers[i-1] and rest > 0:
rest -= 1
elif answers[i] == answers[i-1] and rest == 0:
res += answers[i] + 1
rest = answers[i]
return res参考了讨论里面的答案：写法很优雅速度更慢
class Solution(object):
def numRabbits(self, answers):
"""
:type answers: List[int]
:rtype: int
"""
c = collections.Counter(answers)
return sum((c[i]+i) / (i+1) * (i+1) for i in c)319. Bulb Switcher
思路：
class Solution(object):
def bulbSwitch(self, n):
return int(math.sqrt(n))41. First Missing Positive
做的第一题Hard题，主要难点在O（n）和O（c）space。
思路：不得不说太棒了！
class Solution(object):
def firstMissingPositive(self, nums):
nums.append(0) # because the missing positive interger will be in the range from [1, n+1]
n = len(nums)
for i in xrange(n): # [7, 8, 9, 0] => [0, 0, 0, 0]
if nums[i] < 0 or nums[i] >= n:
nums[i] = 0
for i in xrange(n): # use the index to store the frequence of the num
nums[nums[i]%n] += n # 这里要取余，因为之前可能被加了n
for i in xrange(1, n):
if nums[i] / n == 0:
return i
return n
396. Rotate Function
思路：
F(k) = F(k-1) + Sum - nA(-k)
class Solution(object):
def maxRotateFunction(self, A):
"""
:type A: List[int]
:rtype: int
"""
if not A:
return 0
Sum = sum(A)
n = len(A)
F = [0] * n
for i in xrange(n):
if i == 0:
F[i] = sum(e*i for i, e in enumerate(A))
else:
F[i] = F[i-1] + Sum - n*A[-i]
return max(F) 313. Super Ugly Number
class Solution(object):
def nthSuperUglyNumber(self, n, primes):
"""
:type n: int
:type primes: List[int]
:rtype: int
"""
uglys = [1]
points = [0] * len(primes)
for i in xrange(n - 1):
Min = 2**32
for j in xrange(len(primes)):
Min = min(Min, uglys[points[j]] * primes[j])
for jj in xrange(len(primes)):
if uglys[points[jj]] * primes[jj] == Min:
points[jj] += 1
uglys.append(Min)
return uglys[-1]
334. Increasing Triplet Subsequence
这么简单粗暴的方法为啥我一开始没想出来？
class Solution(object):
def increasingTriplet(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
c1, c2 = 2**31, 2**31
for n in nums:
if n <= c1:
c1 = n
elif n <= c2:
c2 = n
else:
return True
return False50. Pow(x, n)
class Solution(object):
def myPow(self, x, n):
"""
:type x: float
:type n: int
:rtype: float
"""
if n < 0:
return 1 / self.myPow(x, -n)
if n == 0:
return 1
elif n == 2:
return x*x
if n % 2 == 0:
return self.myPow(self.myPow(x, n/2), 2)
else:
return x * self.myPow(self.myPow(x, n/2), 2)43. Multiply Strings
class Solution(object):
def multiply(self, num1, num2):
"""
:type num1: str
:type num2: str
:rtype: str
"""
n1, n2 = len(num1), len(num2)
indices = [0] * ((n1-1) + (n2-1) + 1 + 1)
for j in xrange(n2):
for i in xrange(n1):
num = (ord(num1[i]) - ord('0')) * (ord(num2[j]) - ord('0'))
indices[i+j] += num / 10
indices[i+j+1] += num % 10
for i in xrange(len(indices)-1, 0, -1):
indices[i-1] += indices[i] / 10
indices[i] = str(indices[i] % 10)
indices[0] = str(indices[0])
p = 0
for i in xrange(len(indices)-1):
if indices[i] == '0':
p += 1
else:
break
return "".join(indices[p:])29. Divide Two Integers
class Solution(object):
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
sign = 1
if (dividend > 0 and divisor < 0) or (dividend < 0 and divisor > 0):
sign = -1
dividend, divisor = abs(dividend), abs(divisor)
if divisor == 0:
return sys.maxsize
if dividend == 0 or divisor > dividend:
return 0
res = self._divide(dividend, divisor)
if res > 2**31-1:
return 2**31-1 if sign == 1 else -2**31
return res*sign
def _divide(self, dividend, divisor):
if dividend < divisor:
return 0
_sum = divisor
mutiple = 1
while (_sum + _sum) < dividend:
_sum += _sum # 两倍两倍的加可以提高运行速度
mutiple += mutiple
return mutiple + self._divide(dividend-_sum, divisor)
247. Strobogrammatic Number II
class Solution(object):
def findStrobogrammatic(self, n):
"""
:type n: int
:rtype: List[str]
"""
oddMiddleCandicate = ['0', '1', '8']
evenMiddleCandicate = ["11","69","88","96", '00']
if n == 1:
return oddMiddleCandicate
if n == 2:
return evenMiddleCandicate[:-1]
if n % 2:
pre, middle = self.findStrobogrammatic(n-1), oddMiddleCandicate
else:
pre, middle = self.findStrobogrammatic(n-2), evenMiddleCandicate
mid_idx = (n-1) / 2
return [x[:mid_idx] + y + x[mid_idx:] for x in pre for y in middle]
248. Strobogramm Number III
class Solution(object):
def strobogrammaticInRange(self, low, high):
"""
the low and high numbers are represented as string.
"""
maps = aps={"0":"0","1":"1","6":"9","8":"8","9":"6"}
cl, ch = len(low), len(high) # character length low and high
if cl > ch or (cl == ch and low > high): return 0
ans = ['', '0', '1', '8']
count = 0
while ans:
temp = []
for s in ans:
if ch - len(s) >= 2:
for key in maps:
temp.append(key+s+maps[key])
if cl <= len(s) <= ch: # 过滤掉不满足的，把满足的结果 count+1
if len(s) == cl and s < low: continue
if len(s) == ch and s > high: continue
if len(s) > 1 and s[0] == '0': continue # leading zero
count += 1
ans = temp
return count65. Valid Number
class Solution(object):
def isNumber(self, s):
"""
:type s: str
:rtype: bool
"""
#define a DFA
state = [{},
{'blank': 1, 'sign': 2, 'digit':3, '.':4},
{'digit':3, '.':4},
{'digit':3, '.':5, 'e':6, 'blank':9},
{'digit':5},
{'digit':5, 'e':6, 'blank':9},
{'sign':7, 'digit':8},
{'digit':8},
{'digit':8, 'blank':9},
{'blank':9}]
currentState = 1
for c in s:
if c >= '0' and c <= '9':
c = 'digit'
if c == ' ':
c = 'blank'
if c in ['+', '-']:
c = 'sign'
if c not in state[currentState].keys():
return False
currentState = state[currentState][c]
if currentState not in [3,5,8,9]:
return False
return True