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Commit b6c4471

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1 parent 6bd7dc3 commit b6c4471

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Given a sorted array of integers, find the starting and ending position of a given target value.
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Your algorithm's runtime complexity must be in the order of O(log n).
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If the target is not found in the array, return [-1, -1].
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For example,
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Given [5, 7, 7, 8, 8, 10] and target value 8,
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return [3, 4].
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'''
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class Solution(object):
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def searchRange(self, nums, target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: List[int]
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"""
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for i in range(len(nums)):
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if nums[i] == target:
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left = i
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break
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else:
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return [-1, 1]
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# find the index of the rightmost appearance of `target` (by reverse
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# iteration). it is guaranteed to appear.
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for j in range(len(nums) - 1, -1, -1):
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if nums[j] == target:
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right = j
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break
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return [left, right]
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if __name__ == "__main__":
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assert Solution().searchRange([5, 7, 7, 8, 8, 10], 8) == [3, 4]
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assert Solution().searchRange([5, 7, 7, 8, 8, 10], 5) == [0, 0]
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assert Solution().searchRange([5, 7, 7, 8, 8, 10], 7) == [1, 2]
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assert Solution().searchRange([5, 7, 7, 8, 8, 10], 10) == [5, 5]

‎Python3/038_Count_and_Say.py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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The count-and-say sequence is the sequence of integers beginning as follows:
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1, 11, 21, 1211, 111221, ...
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1 is read off as "one 1" or 11.
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11 is read off as "two 1s" or 21.
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21 is read off as "one 2, then one 1" or 1211.
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Given an integer n, generate the nth sequence.
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Note: The sequence of integers will be represented as a string.
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'''
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class Solution(object):
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def countAndSay(self, n):
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"""
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:type n: int
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:rtype: str
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"""
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result = "1"
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for __ in range(1, n):
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result = self.getNext(result)
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return result
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def getNext(self, s):
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result = []
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start = 0
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while start < len(s):
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curr = start + 1
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while curr < len(s) and s[start] == s[curr]:
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curr += 1
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result.extend((str(curr - start), s[start]))
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start = curr
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return "".join(result)
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if __name__ == "__main__":
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assert Solution().countAndSay(5) == "111221"
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‎Python3/050_Pow(x, n).py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Implement pow(x, n).
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'''
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class Solution(object):
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def myPow(self, x, n):
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"""
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:type x: float
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:type n: int
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:rtype: float
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"""
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flag = 1 if n >= 0 else -1
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result = 1
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n = abs(n)
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while n > 0:
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if n & 1 == 1:
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result *= x
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n >>= 1
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x *= x
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if flag < 0:
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result = 1 / result
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return result
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if __name__ == "__main__":
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assert Solution().myPow(2, 5) == 32
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assert Solution().myPow(2.1, 2) == 4.41

‎Python3/062_Unique_Paths_method1.py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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How many possible unique paths are there?
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Above is a 3 x 7 grid. How many possible unique paths are there?
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Note: m and n will be at most 100.
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'''
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class Solution(object):
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def uniquePaths(self, m, n):
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"""
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:type m: int
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:type n: int
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:rtype: int
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"""
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dp = [[1 for __ in range(n)] for __ in range(m)]
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for i in range(1, n):
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for j in range(1, m):
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dp[j][i] = dp[j - 1][i] + dp[j][i - 1]
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return dp[m - 1][n - 1]
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if __name__ == "__main__":
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assert Solution().uniquePaths(3, 7) == 28

‎Python3/062_Unique_Paths_method2.py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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How many possible unique paths are there?
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Above is a 3 x 7 grid. How many possible unique paths are there?
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Note: m and n will be at most 100.
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'''
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import math
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class Solution(object):
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def uniquePaths(self, m, n):
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"""
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:type m: int
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:type n: int
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:rtype: int
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"""
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m -= 1
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n -= 1
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return int(math.factorial(m + n) / (math.factorial(m) * math.factorial(n)))
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if __name__ == "__main__":
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assert Solution().uniquePaths(3, 7) == 28

‎Python3/069_Sqrt(x).py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Implement int sqrt(int x).
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Compute and return the square root of x.
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'''
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class Solution(object):
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def mySqrt(self, x):
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"""
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:type x: int
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:rtype: int
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"""
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result = 1.0
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while abs(result ** 2 - x) > 0.1:
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result = (result + x / result) / 2
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return int(result)
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if __name__ == "__main__":
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assert Solution().mySqrt(5) == 2
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assert Solution().mySqrt(0) == 0

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