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package class31;
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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.HashSet;
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import java.util.LinkedList;
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import java.util.List;
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import java.util.Queue;
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public class Problem_0127_WordLadder {
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// start,出发的单词
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// to, 目标单位
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// list, 列表
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// to 一定属于list
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// start未必
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// 返回变幻的最短路径长度
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public static int ladderLength1(String start, String to, List<String> list) {
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list.add(start);
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// key : 列表中的单词,每一个单词都会有记录!
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// value : key这个单词,有哪些邻居!
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HashMap<String, ArrayList<String>> nexts = getNexts(list);
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// abc 出发 abc -> abc 0
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//
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// bbc 1
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HashMap<String, Integer> distanceMap = new HashMap<>();
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distanceMap.put(start, 1);
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HashSet<String> set = new HashSet<>();
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set.add(start);
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Queue<String> queue = new LinkedList<>();
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queue.add(start);
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while (!queue.isEmpty()) {
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String cur = queue.poll();
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Integer distance = distanceMap.get(cur);
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for (String next : nexts.get(cur)) {
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if (next.equals(to)) {
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return distance + 1;
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}
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if (!set.contains(next)) {
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set.add(next);
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queue.add(next);
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distanceMap.put(next, distance + 1);
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}
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}
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}
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return 0;
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}
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public static HashMap<String, ArrayList<String>> getNexts(List<String> words) {
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HashSet<String> dict = new HashSet<>(words);
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HashMap<String, ArrayList<String>> nexts = new HashMap<>();
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for (int i = 0; i < words.size(); i++) {
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nexts.put(words.get(i), getNext(words.get(i), dict));
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}
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return nexts;
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}
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// 应该根据具体数据状况决定用什么来找邻居
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// 1)如果字符串长度比较短,字符串数量比较多,以下方法适合
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// 2)如果字符串长度比较长,字符串数量比较少,以下方法不适合
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public static ArrayList<String> getNext(String word, HashSet<String> dict) {
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ArrayList<String> res = new ArrayList<String>();
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char[] chs = word.toCharArray();
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for (int i = 0; i < chs.length; i++) {
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for (char cur = 'a'; cur <= 'z'; cur++) {
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if (chs[i] != cur) {
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char tmp = chs[i];
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chs[i] = cur;
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if (dict.contains(String.valueOf(chs))) {
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res.add(String.valueOf(chs));
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}
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chs[i] = tmp;
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}
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}
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}
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return res;
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}
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public static int ladderLength2(String beginWord, String endWord, List<String> wordList) {
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HashSet<String> dict = new HashSet<>(wordList);
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if (!dict.contains(endWord)) {
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return 0;
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}
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HashSet<String> startSet = new HashSet<>();
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HashSet<String> endSet = new HashSet<>();
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HashSet<String> visit = new HashSet<>();
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startSet.add(beginWord);
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endSet.add(endWord);
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for (int len = 2; !startSet.isEmpty(); len++) {
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// startSet是较小的,endSet是较大的
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HashSet<String> nextSet = new HashSet<>();
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for (String w : startSet) {
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// w -> a(nextSet)
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// a b c
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// 0
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// 1
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// 2
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for (int j = 0; j < w.length(); j++) {
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char[] ch = w.toCharArray();
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for (char c = 'a'; c <= 'z'; c++) {
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if (c != w.charAt(j)) {
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ch[j] = c;
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String next = String.valueOf(ch);
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if (endSet.contains(next)) {
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return len;
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}
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if (dict.contains(next) && !visit.contains(next)) {
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nextSet.add(next);
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visit.add(next);
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}
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}
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}
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}
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}
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// startSet(小) -> nextSet(某个大小) 和 endSet大小来比
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startSet = (nextSet.size() < endSet.size()) ? nextSet : endSet;
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endSet = (startSet == nextSet) ? endSet : nextSet;
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}
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return 0;
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}
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}
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