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You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is 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].
分析：
第一种方式：把words加到hashmap里，然后从i开始，逐一查看是否每个word都在hashmap里，如果存在，则把hashmap中word所对应的count-1，如果hashmap为空，这表示所有的words都已经获取到了，把starting Index加入到list。
但是这种方式过不了。擦。
1 public class Solution { 2 public List<Integer> findSubstring(String s, String[] words) { 3 List<Integer> list = new ArrayList<>(); 4 if (s == null || s.length() == 0 || words == null || words.length == 0) 5 return list; 6 7 int length = words[0].length(); 8 for (int i = 0; i < s.length(); i++) { 9 Map<String, Integer> map = getMap(words); 10 int index = i; 11 boolean allContained = true; 12 while (!map.isEmpty()) { 13 if (index + length > s.length()) { 14 allContained = false; 15 break; 16 } 17 String sub = s.substring(index, index + length); 18 if (map.containsKey(sub)) { 19 map.put(sub, map.get(sub) - 1); 20 if (map.get(sub) == 0) { 21 map.remove(sub); 22 } 23 index += length; 24 } else { 25 allContained = false; 26 break; 27 } 28 } 29 30 if (allContained) { 31 list.add(i); 32 } 33 } 34 return list; 35 } 36 37 private Map<String, Integer> getMap(String[] words) { 38 Map<String, Integer> map = new HashMap<>(); 39 40 for (String str : words) { 41 map.put(str, map.getOrDefault(str, 0) + 1); 42 } 43 return map; 44 } 45 }
还有一种方法，就是先一个window去覆盖S，使得被words里的word都在被覆盖的substring里，那么对于下一个，我们需要把window左边的部分右移一个word出来，这样右边可以继续移动。如果我们当前右边要加入的word已经加够了，或者不存在于words里，我们也需要移动window左边的pointer。很可惜的是，这种方法也过不了，不管了。
1 public List<Integer> findSubstring(String s, String[] words) { 2 Map<String, Integer> all = getMap(words); 3 List<Integer> list = new ArrayList<>(); 4 if (s == null || s.length() == 0 || words == null || words.length == 0) 5 return list; 6 7 int length = words[0].length(); 8 for (int i = 0; i < length; i++) { 9 Map<String, Integer> map = getMap(words); 10 int index = i; 11 int starting = i; 12 while (index + length <= s.length()) { 13 String sub = s.substring(index, index + length); 14 if (map.containsKey(sub)) { 15 map.put(sub, map.get(sub) - 1); 16 if (map.get(sub) == 0) { 17 map.remove(sub); 18 } 19 // find all the words in the hashmap 20 if (map.isEmpty()) { 21 list.add(starting); 22 map.put(s.substring(starting, starting + length), 1); 23 starting = starting + length; 24 } 25 } else { 26 if (all.containsKey(sub)) { 27 map.put(s.substring(starting, starting + length), 28 map.getOrDefault(s.substring(starting, starting + length), 0) + 1); 29 starting = starting + length; 30 continue; 31 } else { 32 map = getMap(words); 33 starting = index + length; 34 } 35 } 36 index += length; 37 } 38 } 39 return list; 40 } 41 42 private Map<String, Integer> getMap(String[] words) { 43 Map<String, Integer> map = new HashMap<>(); 44 45 for (String str : words) { 46 map.put(str, map.getOrDefault(str, 0) + 1); 47 } 48 return map; 49 }
Substring with Concatenation of All Words
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5747889.html
