LeetCode 30. Substring with Concatenation of All Words

最新推荐文章于 2021-03-09 14:57:41 发布

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本文介绍了一个使用 C++ 实现的算法，该算法用于在一个给定的字符串中寻找由多个指定单词组成的子串的所有起始索引。通过使用 STL 中的 unordered_map 来统计和匹配单词出现的次数。

This question is pretty straight forward. Just need map STL.

#include <vector>
#include <string>
#include <unordered_map>
#include <iostream>
using namespace std;

/*
  You are given a string, s and a list of words, that are all of the same length. Find all
  starting indices of substrings in s that a concatenation of each word in words
  exactly once and without any intervening characters.
  For example:
  Given: s "barfoothefoobarman", words ["foo", "bar"]. return indices[0, 9].
*/

vector<int> findSubstring(string s, vector<string>& words) {
  unordered_map<string, int> wordCounts;
  vector<int> res;
  for(int i = 0; i < words.size(); ++i) {
    wordCounts[words[i]]++;
  }
  int N = words.size();
  if(N <= 0) return res;
  int M = words[0].size();
  for(int i = 0; i <= s.size() - N * M; ++i) {
    unordered_map<string, int> currCounts;
    int j = 0;
    for(j = 0; j < N; ++j) {
      string w = s.substr(i + j * M, M);
      if(wordCounts.find(w) == wordCounts.end()) break;
      currCounts[w]++;
      if(currCounts[w] > wordCounts[w]) break;
    }
    if(j == N) res.push_back(i);
  }
  return res;
}

int main(void) {
  string s = "barfoothefoobarman";
  vector<string> words{"foo", "bar"};
  vector<int> index = findSubstring(s, words);
  for(int i = 0; i < index.size(); ++i) {
    cout << index[i] << endl;
  }
}