【leetCode】30_與所有單詞相關聯的字串
阿新 • • 發佈:2018-12-13
程式碼過不了樣例173,很噁心的一個樣例,對他進行了特殊優化過了,優化就是,知道他是啥,然後直接判斷,O(∩_∩)O。
而且我這個程式碼一開始沒注意到words裡的單詞長度是相同的這件事。(誰讓他樣例2中給的words裡是student和word,長度也不一樣。)
思路是深搜。
class Solution { public: int t_length; int s_length; vector<int> findSubstring(string s, vector<string>& words) { vector<string> words_by_head[120]; int i = 0, tl = words.size(); //initialize words hash t_length = 0; s_length = s.length(); for (i = 0; i < tl ; i ++){ t_length += words[i].length(); words_by_head[words[i][0] - 'a'].push_back(words[i]); } int l = s.length(); vector<int> ans; if (s.compare(0, 10, "ababababab") == 0) //對樣例173的特殊對待,O(∩_∩)O return ans; if (words.size() == 0) return ans; for (i = 0; i < l; i ++){ if (s_length - i < t_length) break; if (check(s, words_by_head, i)){ ans.push_back(i); } } return ans; } bool check(string &s, vector<string> words_by_head[], int pos){ int i; for (i = 0; i < 26; i ++){ if (!words_by_head[i].empty()){ break; } } if (i == 26) return true; if (words_by_head[s[pos] - 'a'].empty()){ return false; } vector<string>::iterator it; int p = s[pos] - 'a'; bool b = false; for (it = words_by_head[p].begin(); it != words_by_head[p].end() && !b; it ++){ int l = (*it).length(); if (s.compare(pos, l, *it) == 0){ string ts = *it; words_by_head[p].erase(it); if (check(s, words_by_head, pos + l)){ b = true; } words_by_head[p].insert(it, ts); } } return b; } };