You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in wordsexactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
解题思路:
(1) 怎样得到所有候选解?
一种方法是用大小为words.size()*words[0].size()的window在s上滑动选取候选集,即得到的候选集为:barfoo,arfoot,rfooth... 等等.然后对候选解处理,判断是否是一个concatenation。
另一种方法的思想是:通过words[0].size()次移位覆盖所有候选集,具体思想是:
方法二:引入变量left,假设left = 0,即从第0个元素起开始处理,遍历每个单词,形式为:bar foo the foo bar man; left = 1时,遍历每次单词,为:arf oot hef oob arm; left=2时,遍历每个单词,形式为:rfo oth efo oba rma. 当left>=word[0].size()= 3时,单词的划分形式和left=1相同。因此,上述三种划分覆盖了所有的单词划分形式。而每个候选解即是这些单词的顺序组合。
(2) 怎样判断候选解?
对words建立一个词典,对s建立一个词典,比较s的词典和words中的词典是否吻合,若是,则找到一个符合条件的解。
方法二代码:
class Solution { public: // travel all the words combinations to maintain a window // there are wl(word len) times travel // each time, n/wl words, mostly 2 times travel for each word // one left side of the window, the other right side of the window // so, time complexity O(wl * 2 * N/wl) = O(2N) vector<int> findSubstring(string S, vector<string> &L) { vector<int> ans; int n = S.size(), cnt = L.size(); if (n <= 0 || cnt <= 0) return ans; // init word occurence unordered_map<string, int> dict; for (int i = 0; i < cnt; ++i) dict[L[i]]++; // travel all sub string combinations int wl = L[0].size(); for (int i = 0; i < wl; ++i) { int left = i, count = 0; unordered_map<string, int> tdict; for (int j = i; j <= n - wl; j += wl) { string str = S.substr(j, wl); // a valid word, accumulate results if (dict.count(str)) { tdict[str]++; if (tdict[str] <= dict[str]) count++; else { // a more word, advance the window left side possiablly while (tdict[str] > dict[str]) { string str1 = S.substr(left, wl); tdict[str1]--; if (tdict[str1] < dict[str1]) count--; left += wl; } } // come to a result if (count == cnt) { ans.push_back(left); // advance one word tdict[S.substr(left, wl)]--; count--; left += wl; } } // not a valid word, reset all vars else { tdict.clear(); count = 0; left = j + wl; } } } return ans; } };
class Solution { public: vector<int> findSubstring(string s, vector<string>& words) { vector<int> ans; int sz = s.size(); int n = words.size(); if(!sz || !n) return ans; int l = words[0].size(); int t_size = n*l; unordered_map<string,int> t_map; for(auto c:words){ t_map[c]++; } int start = 0; while(start+t_size<=sz){ unordered_map<string,int> s_map; int count=0; for(int i=start;i<t_size+start;i+=l){ string c = s.substr(i,l); if(t_map[c]>0){ ++s_map[c]; if(s_map[c]==t_map[c]) count+=t_map[c]; else if(s_map[c]>t_map[c]) break; } else break; } if(count==n) ans.push_back(start); start++; } return ans; } };