leetcode----127. Word Ladder

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链接：

https://leetcode.com/problems/word-ladder/

大意：

给定一个单词beginWord以及单词endWord，还有一个词典wordList。要求找出从beginWord转换为endWord的最短序列长度，且每次转换候的单词都必须是wordList中单词，且每次转换只能是两个仅有一位（且是同一位置）不同的字符串进行转换。例子：

思路：

dfs回溯+剪枝。

从endWord往beginWord进行回溯，最终超时...

请看zhazhad代码。

代码：（超时）

class Solution {int minCount = Integer.MAX_VALUE;int curCount = 1; // 初始为1public int ladderLength(String beginWord, String endWord, List<String> wordList) {if (!wordList.contains(endWord))return 0;// 一次转换即可 转换序列长度为2if (oneWordDifferent(beginWord, endWord))return 2;// 从endWord开始dfs逆推表演 将endWord的访问标志置为trueboolean[] visited = new boolean[wordList.size()];for (int i = 0; i < wordList.size(); i++) {if (wordList.get(i).equals(endWord)) {visited[i] = true;break;}}dfs(beginWord, endWord, wordList, visited);return minCount == Integer.MAX_VALUE ? 0 : minCount;}public void dfs(String beginWord, String curWord, List<String> wordList, boolean[] visited) {// System.out.println(beginWord + ":" + curWord + ":" + curCount);if (oneWordDifferent(beginWord, curWord)) {minCount = curCount + 1;return ;}curCount++;int idx = 0;// 剪枝while (curCount < minCount && idx < wordList.size()) {if (!visited[idx] && oneWordDifferent(curWord, wordList.get(idx))) {visited[idx] = true;dfs(beginWord, wordList.get(idx), wordList, visited);visited[idx] = false; // 回溯}idx++;}curCount--;}// 判断两个单词是否只有对应一位不同public boolean oneWordDifferent(String beginWord, String curWord) {int c = 0, idx = 0;while (idx < beginWord.length()) {if (beginWord.charAt(idx) != curWord.charAt(idx))c++;idx++;}return c == 1;}
}

结果：

超时。思考：也许得用BFS。。。

改进：

使用广度优先遍历算法解决。虽然通过了，但是效率好低啊（蠢哭... 急需大神代码安慰

class Solution {public int ladderLength(String beginWord, String endWord, List<String> wordList) {if (!wordList.contains(endWord))return 0;// 一次转换即可 转换序列长度为2if (oneWordDifferent(beginWord, endWord))return 2;boolean[] visited = new boolean[wordList.size()];int count = 1;for (int i = 0; i < wordList.size(); i++) {if (wordList.get(i).equals(endWord)) {visited[i] = true;break;}}List<String> curWords = new ArrayList<>();curWords.add(endWord);while (curWords.size() > 0) {List<String> tmp = new ArrayList<>();for (String s : curWords) {for (int i = 0; i < wordList.size(); i++) {if (!visited[i] && oneWordDifferent(s, wordList.get(i))) {tmp.add(wordList.get(i));visited[i] = true;// 快速判断if (wordList.get(i).equals(beginWord))return count + 1;if (oneWordDifferent(wordList.get(i), beginWord))return count + 2;}}}count += 1;curWords = tmp;}return 0;}// 判断两个单词是否只有对应一位不同public boolean oneWordDifferent(String beginWord, String curWord) {int c = 0, idx = 0;while (idx < beginWord.length()) {if (beginWord.charAt(idx) != curWord.charAt(idx))c++;idx++;}return c == 1;}
}

最佳:

class Solution {public int ladderLength(String beginWord, String endWord, List<String> wordList) {if (beginWord == null || endWord == null || wordList == null) {return 0;}// 将list转为set  查找速度转为O(1)Set<String> dict = new HashSet<>(wordList);if (!dict.contains(endWord)) {return 0;}Set<String> set1 = new HashSet<>();Set<String> set2 = new HashSet<>();set1.add(beginWord);set2.add(endWord);return bfs(set1, set2, dict, 1);}private int bfs(Set<String> set1, Set<String> set2, Set<String> dict, int len) {// 确保每次bfs都是对含元素少的set进行bfsif (set1.size() > set2.size()) {return bfs(set2, set1, dict, len);}Set<String> nextSet = new HashSet<>();for (String word : set1) {char[] chs = word.toCharArray();for (int i = 0; i < chs.length; i++) {char oldChar = chs[i];for (char c = 'a'; c <= 'z'; c++) {// 依次修改chs的每个位置上的字母（改为'a'-'z'） 查看set2是否含有新单词if (c != oldChar) {chs[i] = c;}String newWord = new String(chs);if (set2.contains(newWord)) {return len + 1;}if (dict.contains(newWord)) {nextSet.add(newWord);dict.remove(newWord);}}chs[i] = oldChar; // 将chs[i]修改为原来的字母 下一步修改下一位置的字母}}// nextSet为空表明set1中所有元素都转不成dict中的字符串if (nextSet.isEmpty()) {return 0;}return bfs(nextSet, set2, dict, len + 1);}
}