Leetcode -- Substring with Concatenation of All Words

问题连接：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

问题描述：

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).

问题API： public List<Integer> findSubstring(String S, String[] L)

问题分析：这题其实可以当做一个变种的strstr，strstr是一个字符一个字符的match，这里是不计顺序的连续单词match。因为是不计顺序，这个时候哈希表的运用就变得必要了起来，作为一个特殊的计数器。先构造一个类似字典的HashMap。然后再每个起点创造一个新的HashMap作为一个特殊的匹配计数器，依据这个计数器的内容和当前匹配的内容维护一个窗口，里面全是符合条件的字符串。起点的个数是字符串数组里字符串的长度，然后按照strstr的 延伸思想逐个匹配，匹配到字典并符合条件的，窗口便扩大，一旦窗口的大小等于L
* L[0].length，便表示找到了符合条件的答案之一。一旦当前子字符串不在字典内，当前维护的窗口便彻底作废。具体看代码吧：

public List<Integer> findSubstring(String S, String[] L) {
HashMap<String, Integer> cached_map = new HashMap<String, Integer>();
List<Integer> res = new LinkedList<Integer>();
if(L == null || L.length == 0)
return res;
int head = 0;
for(String str : L)
cached_map.put(str, cached_map.containsKey(str) ? cached_map.get(str) + 1 : 1);
for(int i = 0; i < L[0].length(); i++){
int counter = 0;
head = i;
HashMap<String, Integer> cur_map = new HashMap<String, Integer>();
for(int j = i; j <= S.length() - L[0].length(); j += L[0].length()){
String tmp_str = S.substring(j, j + L[0].length());
if(cached_map.containsKey(tmp_str)){
if(!cur_map.containsKey(tmp_str) || cur_map.get(tmp_str) < cached_map.get(tmp_str)){
counter++;
cur_map.put(tmp_str, cur_map.containsKey(tmp_str) ? cur_map.get(tmp_str) + 1 : 1);
if(counter == L.length){
res.add(head);
String remove = S.substring(head, head + L[0].length());
cur_map.put(remove, cur_map.get(remove) - 1);
head += L[0].length();
counter--;
}
}else{
String remove_candidate = S.substring(head, head + L[0].length());
while(!remove_candidate.equals(tmp_str)){
cur_map.put(tmp_str, cur_map.get(remove_candidate) - 1);
head += L[0].length();
counter--;
remove_candidate = S.substring(head, head + L[0].length());
}
head += L[0].length();
}
}else{
head = j + L[0].length();
counter = 0;
cur_map.clear();
}
}
}
return res;
}