Word Ladder

描述

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.

分析

求最短路径，用广搜。

单队列

// Word Ladder
// 时间复杂度O(n)，空间复杂度O(n)
struct state_t {
    string word;
    int level;

    state_t() { word = ""; level = 0; }
    state_t(const string& word, int level) {
        this->word = word;
        this->level = level;
    }

    bool operator==(const state_t &other) const {
        return this->word == other.word;
    }
};

namespace std {
    template<> struct hash<state_t> {
    public:
        size_t operator()(const state_t& s) const {
            return str_hash(s.word);
        }
    private:
        std::hash<std::string> str_hash;
    };
}


class Solution {
public:
    int ladderLength(const string& start, const string &end,
            const unordered_set<string> &dict) {
        queue<state_t> q;
        unordered_set<state_t> visited;  // 判重

        auto state_is_valid = [&](const state_t& s) {
            return dict.find(s.word) != dict.end() || s.word == end;
        };
        auto state_is_target = [&](const state_t &s) {return s.word == end; };
        auto state_extend = [&](const state_t &s) {
            unordered_set<state_t> result;

            for (size_t i = 0; i < s.word.size(); ++i) {
                state_t new_state(s.word, s.level + 1);
                for (char c = 'a'; c <= 'z'; c++) {
                    // 防止同字母替换
                    if (c == new_state.word[i]) continue;

                    swap(c, new_state.word[i]);

                    if (state_is_valid(new_state) &&
                        visited.find(new_state) == visited.end()) {
                        result.insert(new_state);
                    }
                    swap(c, new_state.word[i]); // 恢复该单词
                }
            }

            return result;
        };

        state_t start_state(start, 0);
        q.push(start_state);
        visited.insert(start_state);
        while (!q.empty()) {
            // 千万不能用 const auto&，pop() 会删除元素，
            // 引用就变成了悬空引用
            const auto state = q.front();
            q.pop();

            if (state_is_target(state)) {
                return state.level + 1;
            }

            const auto& new_states = state_extend(state);
            for (const auto& new_state : new_states) {
                q.push(new_state);
                visited.insert(new_state);
            }
        }
        return 0;
    }
};

双队列

// Word Ladder
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
    int ladderLength(const string& start, const string &end,
            const unordered_set<string> &dict) {
        queue<string> current, next;    // 当前层，下一层
        unordered_set<string> visited;  // 判重

        int level = -1;  // 层次

        auto state_is_valid = [&](const string& s) {
            return dict.find(s) != dict.end() || s == end;
        };
        auto state_is_target = [&](const string &s) {return s == end;};
        auto state_extend = [&](const string &s) {
            unordered_set<string> result;

            for (size_t i = 0; i < s.size(); ++i) {
                string new_word(s);
                for (char c = 'a'; c <= 'z'; c++) {
                    // 防止同字母替换
                    if (c == new_word[i]) continue;

                    swap(c, new_word[i]);

                    if (state_is_valid(new_word) &&
                        visited.find(new_word) == visited.end()) {
                        result.insert(new_word);
                    }
                    swap(c, new_word[i]); // 恢复该单词
                }
            }

            return result;
        };

        current.push(start);
        visited.insert(start);
        while (!current.empty()) {
            ++level;
            while (!current.empty()) {
                // 千万不能用 const auto&，pop() 会删除元素，
                // 引用就变成了悬空引用
                const auto state = current.front();
                current.pop();

                if (state_is_target(state)) {
                    return level + 1;
                }

                const auto& new_states = state_extend(state);
                for (const auto& new_state : new_states) {
                    next.push(new_state);
                    visited.insert(new_state);
                }
            }
            swap(next, current);
        }
        return 0;
    }
};

相关题目

• Word Ladder II