【leetcode刷题笔记】Substring with Concatenation of All Words

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]

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

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题解：题目的意思是在S中找到一个子串，恰好包含了L中所有的串，L中的串在S的字串中的顺序不重要。

思路很简单，假设L中共有m个串，每个串长度为n，那么L中子串合并起来总长度是m*n，那么只要在S中依次搜索长度为m*n的串就可以了。在搜索的过程中，设置两个hashmap，一个存放L中的串和它们在L中出现的次数，一个存放在S中m*n的子串中找到的长度为n的串和它们在S的子串中出现的次数，因为查看的是S长度为m*n的子串，并且是n个字符为一组查看的，所以要么在S中看到某个长度为n的子串不出现在L中，要么在S中出现的次数比L中多，否则这个长度为m*n的串就是L的所有串的合并。

例如题目中的例子

• 我们首先查看S的子串barfoo，查看这个子串的时候，按照bar,foo的顺序查看，得知子串foobar是符合要求的
• 再查看子串arfoot，查看顺序是arf，oot，发现arf不在L中，所以arfoot不符合要求；
• 再查看子串rfooth，......

         if(L == null || L.length == 0)
             return null;
         int m = L.length;
         int n = L[0].length();
         //store n-length strings in L
         HashMap<String, Integer> map = new HashMap<String, Integer>();
         //store n-length strings inS
         HashMap<String, Integer> InS = new HashMap<String, Integer>();
         List<Integer> answer = new ArrayList<Integer>();
         for(String s:L){
             if(!map.containsKey(s))
                 map.put(s, 1);
             else {
                 map.put(s, map.get(s)+1);
             }
         }

         for(int i = 0;i <= S.length() - m*n;i++){
             InS.clear();
             boolean find = true;
             for(int j = 0;j < m;j++){
                 String sub = S.substring(i+j*n,i+(j+1)*n);
                 //if a n-length string in S's substring doesn't in L, skip to search a new substring in S
                 if(!map.containsKey(sub)){
                     find = false;
                     break;
                 }
                 if(!InS.containsKey(sub))
                     InS.put(sub, 1);
                 else {
                         InS.put(sub, InS.get(sub)+1);
                 }
                 //if a n-length string in S'substring appears more time than in L, stop checking this substring
                 if(InS.get(sub) > map.get(sub)){
                     find = false;
                     break;
                 }
             }
             if(find)
                 answer.add(i);
         }
         return answer;