LeetCode 29 Substring with Concatenation of All Words

标签：java   leetcode   

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;
	}
}

LeetCode 29 Substring with Concatenation of All Words

标签：java   leetcode   

原文地址：http://blog.csdn.net/mlweixiao/article/details/39290505