LeetCode-30. Substring with Concatenation of All Words

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 words exactly once and without any intervening characters.

Example 1:

Input:s = "barfoothefoobarman",words = ["foo","bar"]
Output:[0,9]

Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:s = "wordgoodgoodgoodbestword",words = ["word","good","best","word"]
Output:[]

题解：

暴力判断：

class Solution {
public:
  vector<int> findSubstring(string s, vector<string>& words) {
    int n = words.size();
    if (n == 0) {
      return vector<int>();
    }
    int m = words[0].size();
    if (s.length() < n * m) {
      return vector<int>();
    }
    vector<int> res;
    map<string, int> mp;
    for (int i = 0; i < n; i++) {
      if (mp.find(words[i]) == mp.end()) {
        mp.insert(make_pair(words[i], 1));
      }
      else {
        mp[words[i]]++;
      }
    }
    for (int i = 0; i <= s.length() - m * n; i++) {
      int k = i, cnt = 0;
      string sub = s.substr(k, m);
      map<string, int> data(mp);
      while (data[sub] != 0) {
        if (cnt == n - 1) {
          res.push_back(i);
          cnt = 0;
          break;
        }
        data[sub]--;
        k += m;
        sub = s.substr(k, m);
        cnt++;
      }
    }
    return res;
  }
};

滑动窗口：

记words中每个单词长度为m

以0 - m为起点，每次找m个单位大小，这样就能找完所有情况。

如果找到单词并且words[i]中有它并且没用完，cnt++，如果用完了，那就缩小窗口，找到第一个让它空余的最小左边界，如果cnt == n，即该窗口内所有words都出现了并且无多余，存储窗口的左边界。

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        if (s.empty() || words.empty()) {
            return {};
        }
        vector<int> res;
        int l = s.size(), n = words.size(), len = words[0].size();
        unordered_map<string, int> m1;
        for (int i = 0; i < n; i++) {
            m1[words[i]]++;
        }
        for (int i = 0; i < len; ++i) {
            int left = i, cnt = 0;
            unordered_map<string, int> m2;
            for (int j = i; j <= l - len; j += len) {
                string t = s.substr(j, len);
                if (m1.find(t) != m1.end()) {
                    m2[t]++;
                    if (m2[t] <= m1[t]) {
                        cnt++;
                    } else {
                        while (m2[t] > m1[t]) {
                            string tmp = s.substr(left, len);
                            m2[tmp]--;
                            if (m2[tmp] < m1[tmp]) {
                                cnt--;
                            }
                            left += len;
                        }
                    }
                    if (cnt == n) {
                        res.push_back(left);
                        m2[s.substr(left, len)]--;
                        cnt--;
                        left += len;
                    }
                } else {
                    m2.clear();
                    cnt = 0;
                    left = j + len;
                }
            }
        }
        return res;
    }
};