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목록BFS (1)
융융이'Blog
단어 변환
counts = [] def solution(begin, target, words): answer = 0 visited = [0] * len(words) print(visited[0] == 0) if target in words: print("있음") find_array(begin, words, visited, 0, target) answer = print(min(counts)) else: return answer # if last_check(words, target): return answer def check(last, target): count = 0 for i in range(len(target)): if last[i] == target[i]: count += 1 if count == len(ta..
2022이전/알고리즘(하루에하나씩!)
2020. 9. 11. 01:25