126. 单词接龙 II
自己做
解：递归选取（超时）
class Solution {private int min_len = Integer.MAX_VALUE;private List<Boolean> choice; //标记是否选取private List<List<String>> res; //存在结果public void searchWord(String endWord, List<String> wordList, List<Boolean> option, List<String> combine){if(combine.size() > min_len) //比最短组合长也没必要继续往下延伸了return;List<String> combination = new ArrayList(combine); List<Boolean> choice = new ArrayList(option);if(combination.get(combination.size() - 1).equals(endWord)){ //组合中最后一个元素和endWord相同即找到了对应组合if(res.size() == 0){ //第一个组合，直接加入res.add(combination);min_len = combination.size(); }else{ //不是第一个组合，综合考虑if(combination.size() < min_len){ //找到了更短的序列，更新resmin_len = combination.size();//清空以前的结果，以目前最短的为准res = new ArrayList();res.add(combination);}else if(combination.size() == min_len) //找到了同样短的序列，加入resres.add(combination);//比最短组合长的组合就不考虑了}return;}//选取这一层的元素for(int i = 0; i < wordList.size(); i++){if(!choice.get(i)){ //没有被选取过的元素//检查上个单词和当前单词不一样的字母数int num = 0;for(int j = 0; j < combination.get(combination.size() - 1).length(); j++)if(wordList.get(i).charAt(j) != combination.get(combination.size() - 1).charAt(j))num++;if(num == 1){ //相差一个字母，尝试选取combination.add(wordList.get(i));choice.set(i, true);searchWord(endWord, wordList, choice, combination); //放入组合choice.set(i, false);combination.remove(combination.size() - 1);}//除此之外都不选取}}}public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {//检查endWord是否在wordList中boolean is_exist = false;choice = new ArrayList(); for(int i = 0; i < wordList.size(); i++){ //检查的同时顺便初始化choiceif(endWord.equals(wordList.get(i))) //存在is_exist = true;choice.add(false);}if(!is_exist) //不存在就直接返回return new ArrayList();res = new ArrayList();List<String> combination = new ArrayList<String>();combination.add(beginWord);searchWord(endWord, wordList, choice, combination);return res;}
}
看题解
class Solution {public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {List<List<String>> res = new ArrayList<>();// 因为需要快速判断扩展出的单词是否在 wordList 里，因此需要将 wordList 存入哈希表，这里命名为「字典」Set<String> dict = new HashSet<>(wordList);// 特殊用例判断if (!dict.contains(endWord)) {return res;}dict.remove(beginWord);// 第 1 步：广度优先搜索建图// 记录扩展出的单词是在第几次扩展的时候得到的，key：单词，value：在广度优先搜索的第几层Map<String, Integer> steps = new HashMap<String, Integer>();steps.put(beginWord, 0);// 记录了单词是从哪些单词扩展而来，key：单词，value：单词列表，这些单词可以变换到 key ，它们是一对多关系Map<String, List<String>> from = new HashMap<String, List<String>>();int step = 1;boolean found = false;int wordLen = beginWord.length();Queue<String> queue = new ArrayDeque<String>();queue.offer(beginWord);while (!queue.isEmpty()) {int size = queue.size();for (int i = 0; i < size; i++) {String currWord = queue.poll();char[] charArray = currWord.toCharArray();// 将每一位替换成 26 个小写英文字母for (int j = 0; j < wordLen; j++) {char origin = charArray[j];for (char c = 'a'; c <= 'z'; c++) {charArray[j] = c;String nextWord = String.valueOf(charArray);if (steps.containsKey(nextWord) && step == steps.get(nextWord)) {from.get(nextWord).add(currWord);}if (!dict.contains(nextWord)) {continue;}// 如果从一个单词扩展出来的单词以前遍历过，距离一定更远，为了避免搜索到已经遍历到，且距离更远的单词，需要将它从 dict 中删除dict.remove(nextWord);// 这一层扩展出的单词进入队列queue.offer(nextWord);// 记录 nextWord 从 currWord 而来from.putIfAbsent(nextWord, new ArrayList<>());from.get(nextWord).add(currWord);// 记录 nextWord 的 stepsteps.put(nextWord, step);if (nextWord.equals(endWord)) {found = true;}}charArray[j] = origin;}}step++;if (found) {break;}}// 第 2 步：回溯找到所有解，从 endWord 恢复到 beginWord ，所以每次尝试操作 path 列表的头部if (found) {Deque<String> path = new ArrayDeque<>();path.add(endWord);backtrack(from, path, beginWord, endWord, res);}return res;}public void backtrack(Map<String, List<String>> from, Deque<String> path, String beginWord, String cur, List<List<String>> res) {if (cur.equals(beginWord)) {res.add(new ArrayList<>(path));return;}for (String precursor : from.get(cur)) {path.addFirst(precursor);backtrack(from, path, beginWord, precursor, res);path.removeFirst();}}
}、
