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  • leetcode — word-ladder

    import java.util.*;
    
    /**
     * Source : https://oj.leetcode.com/problems/word-ladder/
     *
     *
     * Given two words (start and end), and a dictionary, find the length of shortest
     * transformation sequence from start to end, such that:
     *
     * Only one letter can be changed at a time
     * Each intermediate word must exist in the dictionary
     *
     * For example,
     *
     * Given:
     * start = "hit"
     * end = "cog"
     * dict = ["hot","dot","dog","lot","log"]
     *
     * As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
     * return its length 5.
     *
     * Note:
     *
     * Return 0 if there is no such transformation sequence.
     * All words have the same length.
     * All words contain only lowercase alphabetic characters.
     *
     */
    public class WordLadder {
    
        /**
         *
         * 转化为图的问题
         * start、end、dict中各个单词看做是图中的每个节点
         * 如果有一个单词能变化一个字母到另外一个单词,说明两个节点是连通的
         *
         * 所以就转化为求两个节点之间的最短距离,使用BFS
         *
         * 使用BFS注意:
         * 1. 找到当前需要遍历的节点,这里是当前节点的相邻节点
         * 2. 标记已经遍历过的节点,防止从重复遍历
         *
         * 从start开始使用BFS,遍历当前节点的相邻节点,求出当前节点的相邻节点两种办法:
         * 1. 遍历字典中每个单词,如果和当前单词只差一个字母,说明是相邻的,复杂度为:w*n,w是当前单词的长度,n是字典单词数量
         * 2. 针对当前单词的每个字母,找出可能得变化,每个字母可以变为26个字母中除本身外的其他字母,如果判断变化后的单词在字典中则为相邻的节点,
         *          复杂度为 26*w,w为单词长度
         *
         * 当字典中单词数量较小的时候可以使用第一种方法,如果字典中单词数量较大则使用第二种方法
         *
         * 怎么标记访问过的节点?
         * 已经访问过的节点不需要再次被访问,所以可以从字典中删除
         *
         * 这里使用第二种方法
         *
         * @param start
         * @param end
         * @param dict
         * @return
         */
        public int ladderLength (String start, String end, String[] dict) {
            Set<String> set = new HashSet<String>(Arrays.asList(dict));
            set.add(end);
            Map<String, Integer> map = new HashMap<String, Integer>();
    
            map.put(start, 1);
            while (map.size() > 0) {
                String cur = map.keySet().iterator().next();
                Integer len = map.get(cur);
                if (cur.equals(end)) {
                    System.out.println(len);
    //                return len;
                }
                map.remove(cur);
                Set<String> neighbors = findNeighbors(cur, set);
                for (String str : neighbors) {
                    map.put(str, len+1);
                }
            }
    
            return 0;
    
        }
    
        private Set<String> findNeighbors (String cur, Set<String> dict) {
            Set<String> neighbors = new HashSet<String>();
            for (int i = 0; i < cur.length(); i++) {
                for (int j = 0; j < 26; j++) {
                    char ch = (char) ('a' + j);
                    if (cur.charAt(i) != ch) {
                        String candidate = "";
                        if (i == cur.length()-1) {
                            candidate = cur.substring(0, i) + ch;
                        } else {
                            candidate = cur.substring(0, i) + ch + cur.substring(i+1);
                        }
                        if (dict.contains(candidate)) {
                            neighbors.add(candidate);
                            dict.remove(candidate);
                        }
                    }
                }
            }
            return neighbors;
        }
    
        public static void main(String[] args) {
            WordLadder wordLadder = new WordLadder();
            String start = "hit";
            String end = "cog";
            String[] dict = new String[]{"hot","dot","dog","lot","log"};
            System.out.println(wordLadder.ladderLength(start, end, dict) + "----5");
    
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/sunshine-2015/p/7864814.html
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