LeetCode 29 Substring with Concatenation of All Words

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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;}}

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