LeetCode 30. Substring with Concatenation of All Words（所有单词的连接）

原题网址：https://leetcode.com/problems/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 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).

方法：使用直方图统计。

public class Solution {    public List<Integer> findSubstring(String s, String[] words) {        Map<String, Integer> histograms = new HashMap<>();        for(String word: words) {            Integer count = histograms.get(word);            if (count == null) {                count = 1;            } else {                count ++;            }            histograms.put(word, count);        }        Map<String, Integer> wordCounts = new HashMap<>();        List<Integer> substrings = new ArrayList<>();        int m = words[0].length();        int n = s.length();        for(int offset = 0; offset < m; offset ++) {            int matched = 0;            int pos = offset;            wordCounts.clear();            for(int i=offset; i+m<=n; i+=m) {                String word = s.substring(i, i+m);                if (!histograms.containsKey(word)) {                    wordCounts.clear();                    matched = 0;                    pos = i+m;                    continue;                }                Integer count = wordCounts.get(word);                if (count == null || count < histograms.get(word)) {                    matched ++;                }                if (count == null) {                    count = 1;                } else {                    count ++;                }                wordCounts.put(word, count);                if (matched == words.length) {                    substrings.add(pos);                }                if (pos + (words.length-1)*m <= i) {                    String prev = s.substring(pos, pos+m);                    // System.out.printf("removing %s\n", prev);                    Integer pc = wordCounts.get(prev);                    if (pc <= histograms.get(prev)) matched --;                    pc --;                    wordCounts.put(prev, pc);                    pos += m;                }            }        }        return substrings;    }}

另一种实现：

public class Solution {    public List<Integer> findSubstring(String s, String[] words) {        List<Integer> results = new ArrayList<>();        if (s == null || words == null || words.length == 0) return results;        Map<String, Integer> h = new HashMap<>();        for(String word : words) {            Integer c = h.get(word);            if (c == null) c = 1; else c++;            h.put(word, c);        }        int len = words[0].length();        for(int offset = 0; offset < len; offset++) {            Map<String, Integer> f = new HashMap<>();            int from = offset;            int count = 0;            for(int i = offset; i + len <= s.length(); i += len) {                String word = s.substring(i, i + len);                Integer hc = h.get(word);                if (hc == null) {                    from = i + len;                    count = 0;                    f = new HashMap<>();                    continue;                }                Integer fc = f.get(word);                if (fc == null) fc = 1; else fc++;                f.put(word, fc);                count++;                while (fc > hc) {                    String fword = s.substring(from, from + len);                    f.put(fword, f.get(fword) - 1);                    fc = f.get(word);                    from += len;                    count--;                }                if (count == words.length) {                    results.add(from);                }            }        }        return results;    }}