LeetCode

题目：

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.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

思路：

为words保持一个HashMap

package string;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

public class SubstringWithConcatenationOfAllWords {

    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<Integer>();
        int len;
        int numWords;
        int wordLen;
        if (s == null || (numWords = words.length) == 0 || (len = s.length()) == 0) return res;
        if (words == null || (wordLen = words[0].length()) == 0) return res;
        int totalWordsLen = wordLen*numWords;
        if (totalWordsLen > len) return res;
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        for (String w : words) {
            if (map.containsKey(w)) {
                map.put(w, map.get(w) + 1);
            } else {
                map.put(w, 1);
            }
        }
        
        for (int i = 0; i <= len - totalWordsLen; ++i) {
            if (isSubString(s.substring(i, i + totalWordsLen), new HashMap<String, Integer>(map), wordLen)) {
                res.add(i);
            }
        }
        
        return res;
    }
    
    private boolean isSubString(String subStr, HashMap<String, Integer> map, int wordLen) {
        for (int i = 0; i <= subStr.length() - wordLen; i += wordLen) {
            String word = subStr.substring(i, i + wordLen);
            
            if (map.containsKey(word) && map.get(word) > 0) {
                map.put(word, map.get(word) - 1);
            } else {
                return false; // If !contains or value is 0, return immediately.
            }
        }
        
        for (String word : map.keySet()) {
            if (map.get(word) != 0)
                return false;
        }
        
        return true;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        SubstringWithConcatenationOfAllWords s = new SubstringWithConcatenationOfAllWords();
        String str = "barfoothefoobarman";
        String[] words = {"foo", "bar"};
        for(int i : s.findSubstring(str, words)) {
            System.out.println(i);
        }
    }

}