LeetCode 127. Word Ladder

Solution1：我的超过40%的AC的答案
原先利用BFS做但是内存溢出未能AC；进过修改加上了标记是否访问过的visited数组，可以AC啦~~~

class Solution {
public:int ladderLength(string beginWord, string endWord, vector<string>& wordList) {if (find(wordList.begin(), wordList.end(), endWord) == wordList.end())return 0;queue<string> str_queue;str_queue.push(beginWord);int level = 2, cur = 1, next = 0;vector<int> visited(wordList.size(), 0);while (cur) { string temp = str_queue.front();for (int i = 0; i < wordList.size(); i++) {if (! visited[i] && diff_n_char(temp, wordList[i]) == 1) {visited[i] = 1;if (wordList[i] == endWord)return level;str_queue.push(wordList[i]);next++;}}str_queue.pop();cur--;if (!cur) {cur = next;next = 0;level++;}                }return 0;}private:int diff_n_char(string& s1, string& s2) {int diff = 0;for (int i = 0; i < s1.size(); i++) {if (s1[i] != s2[i])diff++;}return diff;}
};

Solution2：
单向BFS，参考自花花酱：http://zxi.mytechroad.com/blog/searching/127-word-ladder/

class Solution {
public:int ladderLength(string beginWord, string endWord, vector<string>& wordList) {unordered_set<string> dict(wordList.begin(), wordList.end());        if (!dict.count(endWord)) return 0;queue<string> q;q.push(beginWord);int l = beginWord.length();int step = 0;while (!q.empty()) {++step;for (int size = q.size(); size > 0; size--) {                string w = q.front();                q.pop();for (int i = 0; i < l; i++) {                char ch = w[i];for (int j = 'a'; j <= 'z'; j++) {w[i] = j;// Found the solutionif (w == endWord) return step + 1;// Not in dict, skip itif (!dict.count(w)) continue;// Remove new word from dictdict.erase(w);// Add new word into queueq.push(w);                    }w[i] = ch;}}}return 0;}
};//仿写答案
class Solution {
public:int ladderLength(string beginWord, string endWord, vector<string>& wordList) {unordered_set<string> word_set(wordList.begin(), wordList.end());if (!word_set.count(endWord)) //endWord不在wordList中return 0;queue<string> word_q;word_q.push(beginWord);int level = 1, len_word = beginWord.size();while (!word_q.empty()) {level++;for (int i = word_q.size(); i > 0; i--) {//倒着计数很聪明，简化了后面queue增、删元素引起的边界问题string w = word_q.front();word_q.pop();for (int j = 0; j < len_word; j++) {char temp_ch = w[j];for (int k = 'a'; k <= 'z'; k++) {w[j] = k;if (w == endWord) return level;if (!word_set.count(w)) continue;word_set.erase(w);word_q.push(w);}w[j] = temp_ch;}}}return 0;}
};

Solution3：
双向BFS，参考自花花酱：http://zxi.mytechroad.com/blog/searching/127-word-ladder/

class Solution {
public:int ladderLength(string beginWord, string endWord, vector<string>& wordList) {unordered_set<string> dict(wordList.begin(), wordList.end());        if (!dict.count(endWord)) return 0;int l = beginWord.length();unordered_set<string> q1{beginWord};unordered_set<string> q2{endWord};int step = 0;while (!q1.empty() && !q2.empty()) {++step;// Always expend the smaller queue firstif (q1.size() > q2.size())std::swap(q1, q2);unordered_set<string> q;for (string w : q1) {for (int i = 0; i < l; i++) {char ch = w[i];for (int j = 'a'; j <= 'z'; j++) {w[i] = j;if (q2.count(w)) return step + 1;if (!dict.count(w)) continue;                        dict.erase(w);q.insert(w);}w[i] = ch;}}std::swap(q, q1);}return 0;}
};

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