LeetCode – Page 13

Solution to Search a 2D Matrix by LeetCode

17 Jun

Question: https://oj.leetcode.com/problems/search-a-2d-matrix/ Question Name: Search a 2D Matrix

class Solution:

# @param matrix, a list of lists of integers

# @param target, an integer

# @return a boolean

def searchMatrix(self, matrix, target):

''' Flatten the matrix into 1-dimension list. Then this

question turns to be binary search question.

'''

rowLen = len(matrix)

colLen = len(matrix[0])

begin, end = 0, rowLen * colLen - 1

while begin <= end:

mid = (begin + end) // 2

# Convert to 1D binary search question

midCell = matrix[mid//colLen][mid%colLen]

if midCell > target: end = mid - 1

elif midCell < target: begin = mid + 1

else: return True

return False

Solution to Set Matrix Zeroes by LeetCode

17 Jun

Question: https://oj.leetcode.com/problems/set-matrix-zeroes/ Question Name: Set Matrix Zeroes

class Solution:

# @param matrix, a list of lists of integers

# RETURN NOTHING, MODIFY matrix IN PLACE.

def setZeroes(self, matrix):

# Empty matrix

if len(matrix) == 0 or len(matrix[0]) == 0: return

row = [False] * len(matrix)

column = [False] * len(matrix[0])

# Record the rows and columns with element(s) being zero.

for rowIndex in xrange(len(matrix)):

for colIndex in xrange(len(matrix[0])):

if matrix[rowIndex][colIndex] == 0:

row[rowIndex] = True

column[colIndex] = True

# Set the qualified entire row(s) and column(s) to zero.

for rowIndex in xrange(len(matrix)):

for colIndex in xrange(len(matrix[0])):

if row[rowIndex] == True or column[colIndex] == True:

matrix[rowIndex][colIndex] = 0

return

Solution to Edit Distance by LeetCode

16 Jun

Question: https://oj.leetcode.com/problems/edit-distance/ Question Name: Edit Distance

class Solution:

# @return an integer

def minDistance(self, word1, word2):

''' Implement the minimum edit distance according to

the description of WikiPedia:

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

Some other implements are available here:

http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance

'''

delCost = insCost = subCost = 1 # The cost for each operation

steps = range(len(word1)+1)

for word2Index in xrange(len(word2)):

diag = steps[0] # d(i-1, j-1)

steps[0] += 1

for word1Index in xrange(len(word1)):

temp = steps[word1Index+1]

if word1[word1Index] == word2[word2Index]:

steps[word1Index+1] = diag

else:

steps[word1Index+1] = min(steps[word1Index+1] + insCost,

steps[word1Index] + delCost,

diag + subCost)

diag = temp # Store d(i-1, j-1) for next cell

return steps[-1]

Solution to Text Justification by LeetCode

15 Jun

Question: https://oj.leetcode.com/problems/text-justification/ Question Name: Text Justification

class Solution:

# @param words, a list of strings

# @param L, an integer

# @return a list of strings

def fullJustify(self, words, L):

begin, end = 0, 0 # The index of begin word (inclusive) and end

# word (exclusive) in one line

result =[]

while True:

wordsLen = 0 # The total length of all words to be used

# in this line.

# Determine how many following words are used for this line.

while end < len(words):

if wordsLen + len(words[end]) + (end - begin) <= L:

wordsLen += len(words[end])

end += 1

else:

break

else:

# This is the last line.

temp = " ".join(words[begin:end])

temp += " " * (L - len(temp))

result.append(temp)

break

# Construct the line after text justification.

temp = ""

if end == begin + 1:

# This line only contains one word.

temp = words[begin] + " " * (L - len(words[begin]))

else:

# This line contains multiple words.

spaceCount = (L - wordsLen) // (end - begin -1)

oneMoreSpace = (L- wordsLen) % (end- begin - 1)

for index in xrange(begin, end-1):

if index - begin < oneMoreSpace:

temp += words[index] + " " * (spaceCount + 1)

else:

temp += words[index] + " " * spaceCount

temp += words[end-1]

result.append(temp)

begin = end

return result

Solution to Sqrt(x) by LeetCode

15 Jun

Question: https://oj.leetcode.com/problems/sqrtx/ Question Name: Sqrt(x)

class Solution:

# @param x, an integer

# @return an integer

def sqrt(self, x):

begin, end = 0, x

# Use binary search to find the largest

# integer i such that, i * i <= x

while begin <= end:

mid = (begin + end) // 2

temp = mid * mid

if temp == x: return mid

elif temp < x: begin = mid + 1

else: end = mid - 1

return end

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

tl;dr: Please put your code into a <pre>YOUR CODE</pre> section.

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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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