☆☆☆方法1：BFS

☆☆☆☆方法2：雙向BFS，減少搜尋空間大小

程式碼1：BFS

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        if (!wordList.contains(endWord)) {
            return 0;
        }

        Queue<String> queue = new LinkedList<>();
        
 
boolean[] visited = new boolean[wordList.size()];
        queue.offer(beginWord);
        int idx = wordList.indexOf(beginWord);
        if (idx != -1) {
            visited[idx] = true;
        }

        int level = 1; // 初始單詞也算一次
        while (!queue.isEmpty()) {
            int size = queue.size();
            level 
 
++;
            for (int i = 0; i < size; i++) {
                String cur = queue.poll();
                for (int j = 0; j < wordList.size(); j++) {
                    if (visited[j]) continue;
                    String s = wordList.get(j);
                    if (canConvert(cur,s)) {
                        
 
if (s.equals(endWord)) {
                            return level;
                        }
                        queue.offer(s);
                        visited[j] = true;
                    }
                }

            }
        }
        return 0;
    }
    private boolean canConvert(String cur, String s) {
        int diff = 0;
        for (int i = 0; i < cur.length(); i++) {
            if (cur.charAt(i) != s.charAt(i)) {
                diff ++;
                if (diff > 1) break;
            }
        }
        return diff == 1;
    }
}

程式碼2：雙向BFS

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