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Commit 0cd3c5f

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Update to 033
Update to 033
1 parent d5db77b commit 0cd3c5f

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4 files changed

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4 files changed

+164
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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, remove the duplicates in place such that each element appear only once and return the new length.
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Do not allocate extra space for another array, you must do this in place with constant memory.
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For example,
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Given input array nums = [1,1,2],
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Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.
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'''
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class Solution:
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def removeDuplicates(self, nums):
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"""
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:type nums: List[int]
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:rtype: int
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"""
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if not nums:
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return 0
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index = 1
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start = 0
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for i in range(1, len(nums)):
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if nums[start] != nums[i]:
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nums[index] = nums[i]
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index += 1
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start = i
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return index
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if __name__ == "__main__":
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assert Solution().removeDuplicates([1, 1, 2]) == 2
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‎Python3/027_Remove_Element.py‎

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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Given an array and a value, remove all instances of that value in place and return the new length.
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The order of elements can be changed. It doesn't matter what you leave beyond the new length.
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'''
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class Solution:
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def removeElement(self, nums, val):
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"""
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:type nums: List[int]
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:type val: int
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:rtype: int
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"""
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left = 0
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right = len(nums) - 1
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while left <= right:
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while left <= right and nums[left] != val:
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left += 1
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while left <= right and nums[right] == val:
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right -= 1
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if left < right:
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nums[left] = nums[right]
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left += 1
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right -= 1
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return right + 1
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if __name__ == "__main__":
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assert Solution().removeElement([1, 2, 3, 4, 3, 2, 1], 1) == 5
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assert Solution().removeElement([2], 3) == 1
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‎Python3/028_Implement_strStr().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 strStr().
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Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
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Subscribe to see which companies asked this question
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'''
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class Solution:
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def strStr(self, haystack, needle):
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"""
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:type haystack: str
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:type needle: str
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:rtype: int
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"""
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if not needle:
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return 0
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if not haystack:
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return -1
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i = 0
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needleL = len(needle)
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while i < len(haystack):
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a = haystack[i:i + needleL]
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if a == needle:
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return i
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else:
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index = 0
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try:
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index = needle.rindex(haystack[i + needleL])
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except Exception:
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i += needleL + 1
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i += needleL - index
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return -1
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if __name__ == "__main__":
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assert Solution().strStr("abcdefg", "ab") == 0
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assert Solution().strStr("abcdefg", "bc") == 1
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assert Solution().strStr("abcdefg", "cd") == 2
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assert Solution().strStr("abcdefg", "fg") == 5
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assert Solution().strStr("abcdefg", "bcf") == -1
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#!usr/bin/env python3
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# -*- coding:utf-8 -*-
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'''
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Suppose a sorted array is rotated at some pivot unknown to you beforehand.
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(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
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You are given a target value to search. If found in the array return its index, otherwise return -1.
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You may assume no duplicate exists in the array.
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'''
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class Solution(object):
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def search(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: int
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"""
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left = 0
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right = len(nums) - 1
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while left <= right:
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mid = left + (right - left) // 2
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if nums[mid] == target:
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return mid
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if nums[mid] > nums[left]:
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if nums[left] <= target <= nums[mid]:
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right = mid - 1
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else:
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left = mid + 1
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elif nums[mid] < nums[left]:
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if nums[mid] <= target <= nums[right]:
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left = mid + 1
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else:
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right = mid - 1
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else:
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left += 1
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return -1
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if __name__ == "__main__":
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assert Solution().search([4, 5, 6, 7, 0, 1, 2], 4) == 0
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assert Solution().search([4, 5, 6, 7, 0, 1, 2], 7) == 3
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assert Solution().search([4, 5, 6, 7, 0, 1, 2], 0) == 4
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assert Solution().search([4, 5, 6, 7, 0, 1, 2], 2) == 6
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assert Solution().search([3, 1], 3) == 0
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assert Solution().search([3, 1], 1) == 1
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