假设我们有一个称为字典的单词列表,还有另外两个字符串start和end。我们希望通过一次更改一个字符来达到从头到尾的效果,每个产生的单词也应放在字典中。单词区分大小写。因此,我们必须找到最终要达到的最小步骤数。如果不可能,则返回-1。
因此,如果输入是 dictionary = [[“ may” ,“ ray” ,“ rat”] start = “ rat” end = “ may” ,那么输出将是3,因为我们可以选择这条路径: [“ rat” ,“ ray” ,“ may”]。
为了解决这个问题,我们将遵循以下步骤-
dictionary := a new set with all unique elements present in q = a double ended queue with a pair (start, 1) while q is not empty, do (word, distance) := left element of q, and delete the left element if word is same as end, then return distance for i in range 0 to size of word - 1, do for each character c in "abcdefghijklmnopqrstuvwxyz", do next_word := word[from index 0 to i - 1] concatenate c concatenate word[from index (i + 1) to end] if next_word is in dictionary, then delete next_word from dictionary insert (next_word, distance + 1) at the end of q return -1
让我们看下面的实现以更好地理解-
from collections import deque class Solution: def solve(self, dictionary, start, end): dictionary = set(dictionary) q = deque([(start, 1)]) while q: word, distance = q.popleft() if word == end: return distance for i in range(len(word)): for c in "abcdefghijklmnopqrstuvwxyz": next_word = word[:i] + c + word[i + 1 :] if next_word in dictionary: dictionary.remove(next_word) q.append((next_word, distance + 1)) return -1 ob = Solution() dictionary = ["may", "ray", "rat"] start = "rat" end = "may" print(ob.solve(dictionary, start, end))
["may", "ray", "rat"], "rat", "may"
输出结果3