[Leetcode] 17, 127, 126

17. Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

Solution: 对于每一位遍历每种可能性。

Code(递归):

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> ans;        if(digits.size()==0) return ans;        string path;        letterCombinations(0, digits, path, ans);        return ans;    }private:    vector<string> m{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};    void letterCombinations(int index, string& digits, string& path, vector<string>& ans){        if(index==digits.size()){            ans.push_back(path);            return;        }        int number = digits[index]-'0';        for(int i=0; m[number][i]; i++){            path.push_back(m[number][i]);            letterCombinations(index+1, digits, path, ans);            path.pop_back();        }    }};

Code(迭代):

class Solution {public:    const vector<string> m{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};    vector<string> letterCombinations(string digits) {        vector<string> ans;        if(digits.size()==0) return ans;                ans.push_back("");        for(int i=0; i<digits.size(); i++){                        int size = ans.size();            for(int j=0; j<size; j++){                                string s = ans[0];                ans.erase(ans.begin());                                int index = digits[i]-'0';                for(int t=0; m[index][t]; t++){                    s.push_back(m[index][t]);                    ans.push_back(s);                    s.pop_back();                }            }                    }        return ans;    }};

127. Word Ladder

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

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.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution: 广度优先搜索，一层层的搜索相邻单词，使用位向量visited记录访问过的单词去重。

Code:

class Solution {public:    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {        if(beginWord==endWord) return 1;        //广搜        vector<bool> visited;//不要使用vector<bool> visited(false, wordList.size*())的方式初始化，不知为何会出bug        for(int i=0; i<wordList.size(); i++) visited.push_back(false); //直接push_back        queue<pair<int,int>> q;//(wordIndex,step)        for(int t=0; t<wordList.size(); t++){            int count = 0;            for(int j=0; j<wordList[t].size(); j++){                if(beginWord[j]!=wordList[t][j]){                    count++;                    if(count>2) break;                }                                }            if(count==1){                q.push(make_pair(t,2));                visited[t] = true;            }        }        while(!q.empty()){            if(wordList[q.front().first]==endWord){                return q.front().second;            }            for(int t=0; t<wordList.size(); t++){                if(visited[t]) continue;                int count = 0;                for(int j=0; j<wordList[t].size(); j++){                    if(wordList[q.front().first][j]!=wordList[t][j]){                        count++;                        if(count>2) break;                    }                }                if(count==1){                    q.push(make_pair(t,q.front().second+1));                    visited[t] = true;                }            }            q.pop();        }        return 0;    }};

126. Word Ladder II

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWordto endWord, such that:

1. Only one letter can be changed at a time
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

Return

  [    ["hit","hot","dot","dog","cog"],    ["hit","hot","lot","log","cog"]  ]

Note:

• Return an empty list if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution: 这题解法思路跟上一题类似，只是注意三点：1、去重的方式必须要保留所有的最短路径；2、记录下路径的方式；3、退出循环的时机。

Code:

class Solution {public:    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {        vector<vector<string>> ans;        if(beginWord==endWord){            ans.push_back(vector<string> {beginWord});            return ans;        }                //广搜        wordList.insert(wordList.begin(), beginWord);        vector<bool> visited;//不要使用vector<bool> visited(false, wordList.size*())的方式初始化，不知为何会出bug        for(int i=0; i<wordList.size(); i++) visited.push_back(false); //直接push_back        unordered_set<int> cur, next;//(wordIndex,step) //使用set去重        cur.insert(cur.end(), 0);        int level = 1;        bool found = false;        unordered_map<int,vector<int>> father;//记录全部的路径        int target = find(wordList.begin(), wordList.end(), endWord) - wordList.begin();        if(target==wordList.size()) return ans;        int k = 20;        while(!cur.empty() && !found){            for(auto i:cur) visited[i] = true;//防止指向同一层或者上一层的单词            while(!cur.empty()){                cout<<level<<' '<<wordList[*cur.begin()]<<": ";                for(int t=0; t<wordList.size(); t++){                    if(visited[t]) continue;                    int count = 0;                    for(int j=0; j<wordList[t].size(); j++){                        if(wordList[*cur.begin()][j]!=wordList[t][j]){                            count++;                            if(count>2) break;                        }                    }                    if(count==1){                        cout<<wordList[t]<<' ';                        next.insert(next.end(),t);                        father[t].push_back(*cur.begin());                        if(t==target) found = true;                    }                }                cout<<endl;                cur.erase(cur.begin());            }            level++;            swap(cur, next);        }        vector<string> path;        if(found) setPath(father, target, wordList, path, ans);        return ans;    }private:    void setPath(unordered_map<int,vector<int>>& father, int target, vector<string>& wordList,                  vector<string>& path, vector<vector<string>>& ans){        path.insert(path.begin(), wordList[target]);        if(target==0){            ans.push_back(path);        }else{            for(int i=0; i<father[target].size(); i++){                setPath(father, father[target][i], wordList, path, ans);            }        }        path.erase(path.begin());        return;    }};