2014-04-29 04:22
题目：给定一堆长度都相等的单词，和起点、终点两个单词，请从这堆单词中寻找一条变换路径，把起点词变成终点词，要求每次变换只能改一个字母。
解法：Leetcode中有Word Ladder，这题基本思路一致。
代码：
1 // 18.10 Given a list of words, all of same length. Given a source and a destionation words, you have to check if there exists a path between the two. Every time you may change only one letter in the word.
2 // This is my code from leetcode problem set: word ladder
3 #include <string>
4 #include <unordered_map>
5 #include <unordered_set>
6 #include <vector>
7 using namespace std;
8
9 class Solution {
10 public:
11 vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {
12 unordered_map<string, vector<string> > back_trace;
13 vector<unordered_set<string> > level(2);
14
15 dict.insert(start);
16 dict.insert(end);
17
18 int flag, nflag;
19 flag = 0;
20 nflag = !flag;
21 level[flag].insert(start);
22
23 unordered_set<string>::iterator usit;
24 char ch, old_ch;
25 string word;
26 while (true) {
27 flag = !flag;
28 nflag = !nflag;
29 level[flag].clear();
30 for (usit = level[nflag].begin(); usit != level[nflag].end(); ++usit) {
31 dict.erase(*usit);
32 }
33 for (usit = level[nflag].begin(); usit != level[nflag].end(); ++usit) {
34 word = *usit;
35 for (size_t i = 0; i < word.size(); ++i) {
36 old_ch = word[i];
37 for (ch = ‘a‘; ch <= ‘z‘; ++ch) {
38 if (ch == old_ch) {
39 continue;
40 }
41 word[i] = ch;
42 if (dict.find(word) != dict.end()) {
43 back_trace[word].push_back(*usit);
44 level[flag].insert(word);
45 }
46 }
47 word[i] = old_ch;
48 }
49 }
50 if (level[flag].empty() || level[flag].count(end) > 0) {
51 // found or not found
52 break;
53 }
54 }
55
56 single_result.clear();
57 for (size_t i = 0; i < result.size(); ++i) {
58 result[i].clear();
59 }
60 result.clear();
61
62 if (!back_trace.empty()) {
63 recorverPath(back_trace, end);
64 }
65
66 return result;
67 }
68 private:
69 vector<vector<string> > result;
70 vector<string> single_result;
71
72 void recorverPath(unordered_map<string, vector<string> > &back_trace, string cur) {
73 if (back_trace.count(cur) == 0) {
74 // this word has no back trace, it is unreachable.
75 vector<string> single_path(single_result);
76
77 single_path.push_back(cur);
78 reverse(single_path.begin(), single_path.end());
79 result.push_back(single_path);
80 return;
81 }
82
83 const vector<string> &v = back_trace[cur];
84 vector<string>::const_iterator usit;
85
86 single_result.push_back(cur);
87 for (usit = v.begin(); usit != v.end(); ++usit) {
88 recorverPath(back_trace, *usit);
89 }
90 single_result.pop_back();
91 }
92 };
《Cracking the Coding Interview》——第18章：难题——题目10,码迷,mamicode.com
时间： 2024-12-25 20:55:05