LeetCode: Word Ladder解题报告

Word Ladder解题报告

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.

SOLUTION 1:

经典的BFS题目。
想象一下，这个变换过程是一个树，每一层是当前所有的变换结果 ，下一层又是上一层的字符串的所有的变换结果。例子：
HIT
AIT, BIT, CIT, DIT..... 

    HAT, HBT, HCT, HDT.....    HIA, HIB, HIC, HID....

HIT 可以有这么多种变换方式，而AIT, BIT本身也可以以相同的方式展开，这就形成了一个相当大的树。
HIT
AIT, BIT, CIT, DIT.....     HAT, HBT, HCT, HDT.....    HIA, HIB, HIC, HID....
 |    (BIT,CIT这些没有再展开了，画图实在不方便)
 |
AIT, BIT, CIT, DIT...     ABT, ACT, ADT....

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        if (start == null || end == null || dict == null) {
            return 0;
        }
        
        // Bug 1: quese is a interface not a class
        Queue<String> q = new LinkedList<String>();
        q.offer(start);
        
        HashSet<String> set = new HashSet<String>();
        set.add(start);
        
        // Bug 3: lever start from 1;
        int level = 1;
        
        while (!q.isEmpty()) {
            int size = q.size();
            
            level++;
            
            for (int i = 0; i < size; i++) {
                String s = q.poll();
                
                int len = s.length();
                
                for (int j = 0; j < len; j++) {
                    StringBuilder sb = new StringBuilder(s);
                    for (char c = 'a'; c <= 'z'; c++) {
                        // Bug 2: setCharAt
                        sb.setCharAt(j, c);
                        String tmp = sb.toString();
                        
                        // 按照题意，这句应该在前，因为题目并不要求end在dict中。
                        if (tmp.equals(end)) {
                            return level;
                        }
                        
                        // Should be in the dict and not in the hashset.
                        if (set.contains(tmp) || !dict.contains(tmp)) {
                            continue;
                        }
                        
                        set.add(tmp);
                        q.offer(tmp);
                    }
                }
                
            }
        }
        
        // When not found, return 0;
        // "hot", "dog", ["hot","dog"]
        return 0;
    }
}

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/bfs/LadderLength_1218_2014.java