Sheng – Page 21 – Code Says

Solution to Climbing Stairs by LeetCode

15 Jun

Question: https://oj.leetcode.com/problems/climbing-stairs/ Question Name: Climbing Stairs

class Solution:

# @param n, an integer

# @return an integer

def climbStairs(self, n):

''' The number of distinct ways to take n steps is

the Fibonacci number.

http://en.wikipedia.org/wiki/Fibonacci_number

'''

steps = [1, 1]

while len(steps) < n + 1:

steps.append(steps[-1] + steps[-2])

return steps[n]

Solution to Simplify Path by LeetCode

15 Jun

Question: https://oj.leetcode.com/problems/simplify-path/ Question Name: Simplify Path

class Solution:

# @param path, a string

# @return a string

def simplifyPath(self, path):

path = path.split("/")

stack = []

for token in path:

# Useless token. May be duplicate "/" or the leading

# and tailing pseudo token.

if token == "": continue

# Keep in current directory

elif token == ".": continue

# Move to father directory, if there is

elif token == "..":

if len(stack) != 0: stack.pop()

else: continue

# Move the a specific son directory

else: stack.append(token)

return "/" + "/".join(stack)

Solution to Add Binary by LeetCode

15 Jun

Question: https://oj.leetcode.com/problems/add-binary/ Question Name: Add Binary

class Solution:

# @param a, a string

# @param b, a string

# @return a string

def addBinary(self, a, b):

# Make sure a is no longer than b.

if len(a) > len(b): a, b = b, a

aLen, bLen = len(a), len(b)

# Addition for letters with carry

addition = {"0": {"0":{"0":["0", "0"], "1":["1", "0"]},

"1":{"0":["1", "0"], "1":["0", "1"]}},

"1": {"0":{"0":["1", "0"], "1":["0", "1"]},

"1":{"0":["0", "1"], "1":["1", "1"]}}

}

result = ""

# Add a and b[-aLen:]

carry = "0"

for index in xrange(-1, -aLen-1, -1):

temp, carry = addition[a[index]][b[index]][carry]

result = temp + result

# Process the remaining part of b

for index in xrange(-aLen-1, -bLen-1, -1):

temp, carry = addition["0"][b[index]][carry]

result = temp + result

if carry == "0":

# No carry at current digit. Then no change will

# happen to the remaining part of b.

result = b[:index] + result

break

else:

if carry == "1": result = "1" + result

return result

Solution to Valid Number by LeetCode

14 Jun

Question: https://oj.leetcode.com/problems/valid-number/ Question Name: Valid Number

class Solution:

# @param s, a string

# @return, a boolean

def isNumberWithoutE(self, s, decimalAllow = True):

''' Check whether s is a number not in scientific notation

'''

digits = "1234567890"

sLen = len(s)

if sLen == 0: return False

if s[0] == "+" or s[0] == "-":

# Skip the sign

s = s[1:]; sLen -= 1

if sLen == 0: return False

if "." in s:

if sLen == 1 or not decimalAllow: return False

decimalPoint = s.index(".")

for item in s[:decimalPoint] + s[decimalPoint+1:]:

if not item in digits: return False

else:

# All elements are digits (except one decimal point)

return True

else:

for item in s:

if not item in digits: return False

else:

# All elements are digits

return True

# @param s, a string

# @return a boolean

def isNumber(self, s):

''' Check whether s is a valid number.

'''

s = s.strip(" ") # Remove the leading and tailing spaces

if "e" in s:

# Might be a number in scientific notation.

divPos = s.index("e")

return self.isNumberWithoutE(s[:divPos]) and

self.isNumberWithoutE(s[divPos+1:], False)

elif "E" in s:

# Might be a number in scientific notation

divPos = s.index("E")

return self.isNumberWithoutE(s[:divPos]) and

self.isNumberWithoutE(s[divPos+1:], False)

else:

# Might be a number not in scientific notation

return self.isNumberWithoutE(s)

Solution to Plus One by LeetCode

14 Jun

Question: https://oj.leetcode.com/problems/plus-one/ Question Name: Plus One

class Solution:

# @param digits, a list of integer digits

# @return a list of integer digits

def plusOne(self, digits):

digits[-1] += 1 # Plus one to the number

# Normalize the list

carry = 0

for index in xrange(len(digits)-1, -1, -1):

digits[index] += carry

carry = digits[index] // 10

# If no carry in this position, no carry any further.

if carry == 0: break

else: digits[index] %= 10

else:

# There is a carry in the most significant digit

digits.insert(0, 1)

return digits

Solution to Merge Two Sorted Lists by LeetCode

14 Jun

Question: https://oj.leetcode.com/problems/merge-two-sorted-lists/ Question Name: Merge Two Sorted Lists

# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution:

# @param two ListNodes

# @return a ListNode

def mergeTwoLists(self, l1, l2):

head = ListNode(-1) # A pseudo head node

current = head

while l1 != None and l2 != None:

if l1.val <= l2.val:

# Append l1's current node

current.next = l1

l1 = l1.next

current = current.next

else:

# Append l2's current node

current.next = l2

l2 = l2.next

current = current.next

if l1 != None: current.next = l1 # Extend the remaining node(s) in l1

else: current.next = l2 # Extend the remaining node(s) in l2

head = head.next # Remove the pseudo head node

return head

Solution to Unique Paths II by LeetCode

14 Jun

Question: https://oj.leetcode.com/problems/unique-paths-ii/ Question Name: Unique Paths II Similar with these questions: Minimum Path Sum and minimum edit distance.

class Solution:

# @param obstacleGrid, a list of lists of integers

# @return an integer

def uniquePathsWithObstacles(self, obstacleGrid):

# COmpute the dimensions

rowLen = len(obstacleGrid)

colLen = len(obstacleGrid[0])

# Compute the number of possible path in the first column

possiblePathCount = [0] * rowLen

possiblePathCount[0] = 1

# Compute the number of possible path in the remaining column

for colIndex in xrange(colLen):

if obstacleGrid[0][colIndex] == 1:

# This cell is obstacle. Impossible to reach here.

possiblePathCount[0] = 0

for rowIndex in xrange(1, rowLen):

if obstacleGrid[rowIndex][colIndex] == 1:

# This cell is obstacle. Impossible to reach here.

possiblePathCount[rowIndex] = 0

else:

# We may reach here by moving from either left or upper cell.

possiblePathCount[rowIndex] += possiblePathCount[rowIndex-1]

return possiblePathCount[-1]

Solution to Unique Paths by LeetCode

13 Jun

Question: https://oj.leetcode.com/problems/unique-paths/ Question Name: Unique Paths

class Solution:

# @return an integer

def uniquePaths(self, m, n):

# From the top-left cell to bottom-right cell, there are

# totally m+n-2 moves. Among these moves, m-1 is moving

# downwards, and n-1 is moving to right. So the number of

# possible unique paths = C(m+n-2, m-1) = C(m+n-2, n-1).

result = 1

# Because C(m+n-2, m-1) = C(m+n-2, n-1), we are going to

# do one with less steps.

total = m + n - 2

chosen = min(m, n) - 1

# http://en.wikipedia.org/wiki/Binomial_coefficient

for i in xrange(total, total-chosen, -1): result *= i

for i in xrange(chosen): result //= (i+1)

return result

Solution to Minimum Path Sum by LeetCode

12 Jun

Question: https://oj.leetcode.com/problems/minimum-path-sum/ Question Name: Minimum Path Sum

class Solution:

# @param grid, a list of lists of integers

# @return an integer

def minPathSum(self, grid):

# Similar with the minimum edit distance algorithm

colLen = len(grid)

column = [-1] * colLen

# Go through the first column

column[0] = grid[0][0]

for index in xrange(1, colLen):

column[index] = grid[index][0] + column[index-1]

# Go through the remaining cells column by column

for colIndex in xrange(1, len(grid[0])):

column[0] += grid[0][colIndex]

for rowIndex in xrange(1, colLen):

column[rowIndex] = min(column[rowIndex], column[rowIndex-1])

+ grid[rowIndex][colIndex]

return column[-1]

Solution to Insert Interval by LeetCode

12 Jun

Question: https://oj.leetcode.com/problems/insert-interval/ Question Name: Insert Interval

# Definition for an interval.

# class Interval:

# def __init__(self, s=0, e=0):

# self.start = s

# self.end = e

class Solution:

# @param intervals, a list of Intervals

# @param newInterval, a Interval

# @return a list of Interval

def insert(self, intervals, newInterval):

if len(intervals) == 0: return [newInterval] # Empty input

# Insert the newInterval according to its start position.

# Actually, we could only find out where to insert, and then

# divide the following merge code into three parts to avoid insertion.

# But the non-insert version would be long code.

# Find the insert position with binary method.

begin, end = 0, len(intervals) - 1

while begin <= end:

mid = (begin + end) // 2

if newInterval.start > intervals[mid].start:

begin = mid + 1

elif newInterval.start < intervals[mid].start:

end = mid - 1

else:

begin = mid

break

# Insert the new interval into the old list.

intervals = intervals[:begin] + [newInterval] + intervals[begin:]

# Do merge if we could

previous = Interval(intervals[0].start, intervals[0].end)

result = []

for interval in intervals[1:]:

if interval.start > previous.end:

# Cannot merge the current interval with the previous one

result.append(previous)

previous = Interval(interval.start, interval.end)

else:

# Merge the current interval with the previous one, by

# extending the end position if possible

previous.end = max(interval.end, previous.end)

# Record the last one interval

result.append(previous)

return result

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If you had some troubles in debugging your solution, please try to ask for help on StackOverflow, instead of here.

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1. To post your code, please add the code inside a <pre> </pre> section (preferred), or <code> </code>. And inside the pre or code section, you do not need to escape < > and &, e.g. no need to use < instead of <.

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