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#yyds干货盘点# leetcode算法题：串联所有单词的子串
source link: https://blog.51cto.com/u_13321676/5396944
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#yyds干货盘点# leetcode算法题：串联所有单词的子串
原创给定一个字符串 s 和一些 长度相同 的单词 words 。找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。
注意子串要与 words 中的单词完全匹配，中间不能有其他字符 ，但不需要考虑 words 中单词串联的顺序。
输入：s = "barfoothefoobarman", words = ["foo","bar"]
输出：[0,9]
从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。
输入：s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
输出：[]
输入：s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
输出：[6,9,12]
代码实现：
class Solution {
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0) return res;
HashMap<String, Integer> map = new HashMap<>();
int one_word = words[0].length();
int word_num = words.length;
int all_len = one_word * word_num;
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
for (int i = 0; i < s.length() - all_len + 1; i++) {
String tmp = s.substring(i, i + all_len);
HashMap<String, Integer> tmp_map = new HashMap<>();
for (int j = 0; j < all_len; j += one_word) {
String w = tmp.substring(j, j + one_word);
tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
}
if (map.equals(tmp_map)) res.add(i);
}
return res;
}
}
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0) return res;
HashMap<String, Integer> map = new HashMap<>();
int one_word = words[0].length();
int word_num = words.length;
int all_len = one_word * word_num;
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
for (int i = 0; i < s.length() - all_len + 1; i++) {
String tmp = s.substring(i, i + all_len);
HashMap<String, Integer> tmp_map = new HashMap<>();
for (int j = 0; j < all_len; j += one_word) {
String w = tmp.substring(j, j + one_word);
tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
}
if (map.equals(tmp_map)) res.add(i);
}
return res;
}
}
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