Leetcode--Substring with concatenation of all Words

Title Description:
You is given a string, s, and a list of words, words, that is all of the same length. Find all starting indices of substring (s) in S that's 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).


The title means that given 1 strings s and a word array words, the number is n, and the length of each word is L.
Scan S, judging within the L * N range, determine if each word in words appears only once in S [I, l*n] (if words contains 1 words multiple times, where I∈[0,len (s)-l*n)
Note that if the words itself contains duplicate words, then this word also allows the same number of occurrences in S[i, L*n]




Ideas:
1. Use the hash table 1:hash, traverse the word array, save all the words and the number of occurrences
2. Using hash table 2:HASH2, intercept the string str of length L in S, if Str appears in Hash1, add to Hash2, where l=s[i, Words[j] * l]
3. Each time you determine whether the element in the HASH1 is exactly equal to HASH2


Implementation code:

public class Solution {public    ilist<int> findsubstring (string s, string[] words) {           if (string. Isnullorwhitespace (s) | | Words = = NULL | | S.length < words. Length * Words[0]. Length) {return new list<int> ();} var Wlen = words[0]. Length;var result = new list<int> (); var len = words. Length * Wlen;var hash = new dictionary<string, int> (); for (var i = 0;i < words. Length; i++) {if (!hash. ContainsKey (Words[i])) {hash. ADD (Words[i], 1);} Else{hash[words[i]] + +;}} var h = new dictionary<string, int> (), for (var i = 0; i < S.length-len + 1; i++) {for (var j = 0;j < words. Length; J + +) {var str = s.substring (i + j * Wlen, Wlen), if (hash. ContainsKey (str)) {if (!h.containskey (str)) {h.add (str,1);} ELSE{H[STR] + +;}}} if (hash. All (X=>h.containskey (X.key) && h[x.key] = = X.value)) {result. ADD (i);} var k2 = new list<string> (H.keys); foreach (var k in K2) {h[k] = 0;}} return result;}    }

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Leetcode--Substring with concatenation of all Words