[leetcod]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).

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

思路1：假設L中的單位長度為n，依次從S中取長度為n的子串，如果在L中，就記下來。需要藉助hash或map，如果整個L都匹配完了，就算是一個concatenation；當匹配錯誤的時候，S右移一個位置。

思路2（參考2）：最佳化，雙指標法，思想類似 Longest Substring Without Repeating Characters， 複雜度可以達到線性。

 

思路1：

public class Solution {public ArrayList<Integer> findSubstring(String S, String[] L) {if (S == null || L == null)return null;int size = L.length;int len = L[0].length();ArrayList<Integer> res = new ArrayList<Integer>();HashMap<String, Integer> expected = new HashMap<String, Integer>();for (String each : L) {Integer old = expected.get(each);if (old == null)expected.put(each, 1);elseexpected.put(each, old + 1);}HashMap<String, Integer> real = new HashMap<String, Integer>();int i;for (i = 0; i <= S.length() - size * len; i++) {real.clear();int j, k = 0;for (j = i; j < i + size * len; j = j + len, k++) {String sub = S.substring(j, j + len);if (expected.containsKey(sub)) {Integer old = real.get(sub);if (old == null)real.put(sub, 1);elsereal.put(sub, old + 1);if (real.get(sub) > expected.get(sub))break;} elsebreak;}if (k == size)res.add(i);}return res;}public static void main(String[] args) {// String S = "barfoothefoobarman";// String[] L = { "foo", "foo" };// System.out.println(new Solution().findSubstring(S, L));String S = "a";String[] L = { "a" };System.out.println(new Solution().findSubstring(S, L));}}

思路2（待實現中）：

 

 

參考：

http://blog.csdn.net/ojshilu/article/details/22212703

http://blog.csdn.net/linhuanmars/article/details/20342851