LeetCode 29 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).

思路：

第一步：对数组L进行处理，建立一个hashmap,为后面的处理做准备，不能建立hashset，因为L中有可能有重复的，也不能建立ArrayList或者LinkedList，处理大数据会超时。

第二步：若从S[a]~S[b]中包含L中所有字串，我们在处理S[a+L[0].length()] ~ S[b+L[0].length()] 时只需要判断S[b]~S[b+L[0].length()]与S[a]~S[a+L[0].length()]相等即可，若相等，则说明S[a+L[0].length()]
~ S[b+L[0].length()]也是符合要求的，若不相等，则不符合要求，直接否却掉。

public List<Integer> findSubstring(String S, String[] L) {
		ArrayList<Integer> result = new ArrayList<Integer>();
		int len = L.length * L[0].length();
		HashMap<String, Integer> hm = new HashMap<String, Integer>();
		for (int i = 0; i < L.length; i++) {
			if (hm.containsKey(L[i])) {
				int value = hm.get(L[i]);
				value++;
				hm.put(L[i], value);
			} else {
				hm.put(L[i], 1);
			}
		}

		for (int i = 0; i <= S.length() - L[0].length() * L.length; i++) {
			HashMap<String, Integer> hashmap = new HashMap<String, Integer>(hm);
			String str = S.substring(i+len-L[0].length(), i + len);
			if (result.contains(i - L[0].length())) {
				String temp = S.substring(i - L[0].length(), i);
				if (temp.contains(str)) {
					result.add(i);
				}
			} else {
				boolean flag = true;
				for (int j = 0; j < len; j += L[0].length()) {
					str = S.substring(i + j, i + j + L[0].length());
					if (!hashmap.containsKey(str) || hashmap.get(str) == 0) {
						flag = false;
						break;
					} else {
						int value = hashmap.get(str);
						value--;
						hashmap.put(str, value);
					}
				}
				if (flag)
					result.add(i);
			}
		}

		return result;
	}
}

若某个词不属于L中的，则只要包含该词的序列都不用考虑，例如S=aabbccaabb, L[]={aa,bb} 则包含bc的序列abbc和bcca是不行的，包含CC的序列bbcc和ccaa是不行的，包含ca的序列bbca，caab也是不行的，则这些序列均不用考虑，优化后的代码：

public class Solution {
	public List<Integer> findSubstring(String S, String[] L) {
		ArrayList<Integer> result = new ArrayList<Integer>();
		int len = L.length * L[0].length();
		HashMap<String, Integer> hm = new HashMap<String, Integer>();
		for (int i = 0; i < L.length; i++) {
			if (hm.containsKey(L[i])) {
				int value = hm.get(L[i]);
				value++;
				hm.put(L[i], value);
			} else {
				hm.put(L[i], 1);
			}
		}

		HashSet<Integer> forbidden = new HashSet<Integer>();
		for (int i = 0; i <= S.length() - L[0].length() * L.length; i++) {
			if (forbidden.contains(i))
				continue;

			HashMap<String, Integer> hashmap = new HashMap<String, Integer>(hm);
			String str = S.substring(i + len - L[0].length(), i + len);
			if (result.contains(i - L[0].length())) {
				String temp = S.substring(i - L[0].length(), i);
				if (temp.contains(str)) {
					result.add(i);
				}
			} else {
				boolean flag = true;
				for (int j = 0; j < len; j += L[0].length()) {
					str = S.substring(i + j, i + j + L[0].length());
					if (!hashmap.containsKey(str)) {
						for (int k = i + L[0].length(); k <= i + j; k += L[0]
								.length()) {
							forbidden.add(k);
						}
						flag = false;
						break;
					} else if (hashmap.get(str) == 0) {
						flag = false;
						break;
					} else {
						int value = hashmap.get(str);
						value--;
						hashmap.put(str, value);
					}
				}
				if (flag)
					result.add(i);
			}
		}

		return result;
	}
}