[LeetCode-JAVA] Substring with Concatenation of All Words

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 wordsexactly once and without any intervening characters.

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

You should return the indices: [0,9].

题意：题意很重要，开始理解错了，以为要求的是words中每个单词出现的index，真正的题意是，words中每一个单词都出现一次，并且中间不夹杂着其他的单词，即word中所有字符串组成的所有可能在s中出现的index。

思路：建立一个map来维护words的内容，key为words的string，value为对应的数量。 由于words的单词长度相同，因此每次查找的窗口大小一致，为words所有单词的总长度，不用判断边界条件。在判断过程中，建立一个临时的map，来与初始的map进行比较，符合的时候，向list中加入此时的index

代码：

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> list = new ArrayList<Integer>();
        if(s == null || s.length() == 0 || words.length == 0)
            return list;
        
        int wordCount = words.length;
        int wordlen = words[0].length();
        int len = wordlen * wordCount;  //总长度
        
        Map<String, Integer> map = new HashMap<String, Integer>();
        Map<String, Integer> cur = new HashMap<String, Integer>();
        
        for(int i = 0 ; i < wordCount ; i++){
            if(!map.containsKey(words[i]))
                map.put(words[i], 1);
            else
                map.put(words[i], map.get(words[i])+1);
        }
        boolean flag = true; //标志位
        for(int i = 0 ; i < s.length() - len + 1 ; i++){
            cur.clear(); //循环开始的时候清空上一次的操作
            flag = true;
            for(int j = 0 ; j < wordCount ; j++){
                String temp = s.substring(i + j*wordlen, i + wordlen*(j+1)); //每一个窗口
                if(!map.containsKey(temp)){
                    flag = false;
                    break;
                }
                    
                if(!cur.containsKey(temp))
                    cur.put(temp, 1);
                else
                    cur.put(temp, cur.get(temp)+1);
                
                if(cur.get(temp) > map.get(temp)){
                    flag = false;
                    break;
                }
            }
            if(flag) // 如果不是意外break 则内容相符
                list.add(i);
        }
        return list;
    }
}

上面的是自己的思路，不是最快速的，大概是700ms 下面将内外循环换一下 速度提升了一倍 300ms 

代码：

public class Solution {
    public static ArrayList<Integer> findSubstring(String S, String[] L) { 
         ArrayList<Integer> res = new ArrayList<Integer>();
         if(S==null||L==null||S.length()==0||L.length==0)
            return res;
         int wordLen = L[0].length();//same length for each word in dictionary
         
         //put given dictionary into hashmap with each word's count
         HashMap<String, Integer> dict = new HashMap<String, Integer>();
         for(String word: L){
             if(!dict.containsKey(word))
                dict.put(word, 1);
             else
                dict.put(word, dict.get(word) + 1);
         }
         
         for(int i = 0; i < wordLen; i++){
             int count = 0;
             int index = i;//index of each startpoint
             HashMap<String, Integer> curdict = new HashMap<String, Integer>();
             //till the first letter of last word 
             for(int j = i; j <= S.length() - wordLen; j += wordLen){
                 String curWord = S.substring(j, j + wordLen);
                 //check each word to tell if it existes in give dictionary
                 if(!dict.containsKey(curWord)){
                     curdict.clear();
                     count = 0;
                     index = j + wordLen;
                 }else{
                     //form current dictionary
                     if(!curdict.containsKey(curWord))
                        curdict.put(curWord, 1);
                     else
                        curdict.put(curWord, curdict.get(curWord) + 1);
                     
                     //count for current found word and check if it exceed given word count
                     if(curdict.get(curWord) <= dict.get(curWord)){
                         count++;
                     }else{
                         while(curdict.get(curWord) > dict.get(curWord)){
                            String temp = S.substring(index, index + wordLen);
                            curdict.put(temp, curdict.get(temp)-1);
                            if(curdict.get(temp)<dict.get(temp)){
                                count--;
                            }
                            index = index + wordLen;
                        }
                     }
                     
                     //put into res and move index point to nextword 
                     //and update current dictionary as well as count num
                     if(count == L.length){
                         res.add(index);
                         String temp = S.substring(index, index + wordLen);
                         curdict.put(temp, curdict.get(temp)-1);
                         index = index + wordLen;
                         count--;
                     }
                 }
             }//end for j
         }//end for i
          return res;
        }
}

参考链接：http://www.cnblogs.com/springfor/p/3872516.html