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80 lines
2.3 KiB
Python
80 lines
2.3 KiB
Python
from typing import List
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class AcceptedMedium:
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def reverse_int(self, x: int) -> int:
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"""
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Link: https://leetcode.com/problems/reverse-integer/
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Runtime: 36 ms
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"""
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ctr = 1
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if x < 0:
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ctr = -1
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x *= -1
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nr = 0
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while x > 0:
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nr = (nr + (x % 10)) * 10
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x //= 10
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nr //= 10
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return nr * ctr if -2147483648 <= nr <= 2147483647 else 0
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def letter_combinations(self, digits: str) -> List[str]:
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"""
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Link: https://leetcode.com/problems/letter-combinations-of-a-phone-number/
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Runtime: 39 ms
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"""
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# sourcery skip: use-itertools-product
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number_to_string = {'2': ['a', 'b', 'c'], '3': ['d', 'e', 'f'], '4': ['g', 'h', 'i'],
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'5': ['j', 'k', 'l'], '6': ['m', 'n', 'o'], '7': ['p', 'q', 'r', 's'],
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'8': ['t', 'u', 'v'], '9': ['w', 'x', 'y', 'z']}
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l = ['']
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l2 = []
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for n in digits:
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for s in number_to_string[n]:
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for i in l:
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l2.append(i+s)
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l = l2.copy()
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l2 = []
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return [] if l[0] == '' else l
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def convert_linear_str_to_zigzag(s: str, numRows: int) -> str:
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# sourcery skip: use-contextlib-suppress
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"""
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Link: https://leetcode.com/problems/zigzag-conversion/
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Runtime: 72 ms
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"""
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len_s = len(s)
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if len_s < 3 or numRows == 1: return s
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def x(l, t):
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if len(l) < t:
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l.append(None)
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x(l, t)
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return l
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l = []
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t = 0
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while t < len_s:
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l.append(list(s[t:numRows+t]))
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t += numRows
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l.append(list(reversed(x(list(s[t:numRows-2+t]), numRows-2))))
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t += numRows-2
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l = l[:-1] if l[-1] == [] else l
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len_l = len(l)
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s1 = ''
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s2 = ''
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s3 = ''
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t = len(l[0]) - 2
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for i in range(0, len_l, 2):
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s1 += l[i].pop(0)
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try:s3 += l[i].pop(t)
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except IndexError: pass
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for i in range(len(l[0])):
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for j in range(len_l):
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try:
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if l[j][i] is not None:
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s2 += l[j][i]
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except IndexError: pass
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return s1 + s2 + s3 |