Leetcode Substring with Concatenation of All Words
You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
题目意思: 输入字符串S和列表L,L含有一组长度相同的单词。找出所有子串的开始下标,该子串由L的所有单词拼接而成,而且没有夹杂其他字符
解题思路是:
从原串S的第一个字符开始,取长度为L的元素个数乘以L里单词的长度,然后判断该串是否仅仅含了L得所有单词。
将子串拆分成多个长度和L单词长度一致的单词,然后根据hash_map来匹配子串的单词
如果单词匹配成功,则匹配数加1;
如果最后的匹配数等同于L的元素的个数,那么该串包含了L得所有单词,而且把该串的开始位置放入结果集中,继续考察S的下一个字符开始的字串情况。
下面代码不知为什么会超时
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: vector<int> findSubstring(string S, vector<string> &L) { vector<int> res; int l_len = L.size(), sub_l_len = L[0].length(); int subLen = l_len * sub_l_len; unordered_map<string, int> hash_map; for(int i = 0 ; i < l_len; ++ i ) hash_map[L[i]]++; for(int i = 0 ; i < S.length()-subLen+1; ++i){ string sub = S.substr(i,subLen); int cnt = 0; unordered_map<string,int> cpHashMap(hash_map); for(int j = 0; j < l_len; ++ j){ string word = sub.substr(j*sub_l_len,sub_l_len); if(cpHashMap.find(word)==cpHashMap.end() || cpHashMap[word] == 0) break; else{ cpHashMap[word]--; cnt++; } } if(cnt == l_len) res.push_back(i); } return res; } };
对上面代码进行优化,减少了hash_map的拷贝
class Solution { public: vector<int> findSubstring(string S, vector<string> &L) { vector<int> res; int l_len = L.size(), sub_l_len = L[0].length(); int subLen = l_len * sub_l_len; unordered_map<string, int> hash_map,curHashMap; for(int i = 0 ; i < l_len; ++ i ) hash_map[L[i]]++; for(int i = 0 ; i < S.length()-subLen+1; ++i){ string sub = S.substr(i,subLen); int cnt = 0; curHashMap.clear(); for(int j = 0; j < l_len; ++ j){ string word = sub.substr(j*sub_l_len,sub_l_len); if(hash_map.find(word) ==hash_map.end()) break; curHashMap[word]++; if(curHashMap[word] > hash_map[word]) break; cnt++; } if(cnt == l_len) res.push_back(i); } return res; } };