方法一:双向广搜
class Solution { public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { set<string> se; for(auto it : wordList) { se.insert(it); } if(!se.count(endWord)) return 0; auto update = [&](queue<string> &q, unordered_map<string, int> &cur, unordered_map<string, int> &other) -> int { int m = q.size(); while(m -- ) { string t = q.front(); q.pop(); int n = t.size(); for(int i = 0; i < n; i ++ ) { for(int j = 0; j < 26; j ++ ) { string sub = t; sub[i] = 'a' + j; if(se.count(sub)) { if(cur.count(sub)) continue; if(other.count(sub)) { return cur[t] + other[sub] + 1; } else { q.push(sub); cur[sub] = cur[t] + 1; } } } } } return -1; }; auto bfs = [&]() -> int { queue<string> q1, q2; unordered_map<string, int> m1, m2; q1.push(beginWord); m1[beginWord] = 0; q2.push(endWord); m2[endWord] = 0; while(q1.size() && q2.size()) { int t = -1; if(q1.size() <= q2.size()) { //注意这里等于号不能少,确保第一次若两个相等从起点开始搜索 t = update(q1, m1, m2); } else { t = update(q2, m2, m1); } if(t != -1) return t; } return -1; }; int res = bfs(); return res == -1 ? 0: res + 1; } };
标签:q1,sub,--,cur,int,接龙,endWord,LeetCode,size From: https://www.cnblogs.com/zk6696/p/17615964.html