import java.util.*;
/**
* Source : https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/
*
* Created by lverpeng on 2017/7/13.
*
* 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).
*
*
* 找出L中单词连接成的子串在字符串S中出现所有位置
*
*/
public class SubstringWithConcatenationOfAllWords {
/**
*
* 将strArr中的所有单词放在hash表中，单词为key，相同单词出现的次数为value
*
* strArr中所有单词长度一致，S重要出现所有单词的连接成为的字符串，那么S的长度一定要大于strArr中所有单词的总长度，
* 也就是起始字符在0-(S.length - strArr.length * strArr[0].length)之间，
* 以上面的位置为起始位置，依次判断接下来的strArr.length个单词是否正好是上面hash表中的单词，并判断相同单词出现次数，
* 如果没有出现则退出循环，如果出现的次数大于hash表中单词的次数也break
* 一轮循环完成之后判断循环的次数是否正好是strArr.length，如果是，说明S包含strArr连接的字符串，记录此时的起始位置到结果中
*
* @param S
* @param strArr
* @return
*/
public int[] findSubstring (String S, String[] strArr) {
if (S.length() < 1 | strArr.length < 1) {
return new int[]{};
}
Map<String, Integer> wordMap = new HashMap<String, Integer>(); // 存放strArr单词的哈希表
for (String str : strArr) {
if (wordMap.keySet().contains(str)) {
wordMap.put(str, wordMap.get(str) + 1);
} else {
wordMap.put(str, 1);
}
}
int loopCount = 0;
int arrLen = strArr.length;
int wordLen = strArr[0].length();
int arrStrLen = arrLen * wordLen;
List<Integer> result = new ArrayList<Integer>();
for (int i = 0; i < S.length() - arrStrLen; i++) {
int j = 0;
Map<String, Integer> subStrMap = new HashMap<String, Integer>();
for (j = 0; j < arrLen; j++) {
loopCount ++;
String subStr = S.substring(i + j * wordLen, i + j * wordLen + wordLen);
if (!wordMap.keySet().contains(subStr)) {
break;
} else {
if (subStrMap.keySet().contains(subStr)) {
subStrMap.put(subStr, subStrMap.get(subStr) + 1);
} else {
subStrMap.put(subStr, 1);
}
}
if (subStrMap.get(subStr) > wordMap.get(subStr)) {
break;
}
}
if (j == arrLen) {
result.add(i);
}
}
System.out.println("loopCount------->" + loopCount);
int[] res = new int[result.size()];
for (int i = 0; i < result.size(); i++) {
res[i] = result.get(i);
}
return res;
}
/**
* 上面是以步长为1进行循环，下面以步长为word长度进行循环
*
* @param S
* @param strArr
* @return
*/
public int[] findSubstring1 (String S, String[] strArr) {
if (S.length() < 1 || strArr.length < 1) {
return new int[]{};
}
Map<String, Integer> wordMap = new HashMap<String, Integer>();
for (int i = 0; i < strArr.length; i++) {
if (wordMap.keySet().contains(strArr[i])) {
wordMap.put(strArr[i], wordMap.get(strArr[i]));
} else {
wordMap.put(strArr[i], 1);
}
}
List<Integer> result = new ArrayList<Integer>();
int wordLen = strArr[0].length();
Map<String, Integer> subStrMap = new HashMap<String, Integer>();
int loopCount = 0;
for (int i = 0; i < wordLen; i++) {
int count = 0;
int left = i; // 记录待匹配子串起始位置
for (int j = i; j < S.length() - wordLen; j += wordLen) {
loopCount ++;
String subStr = S.substring(j, j + wordLen);
if (wordMap.keySet().contains(subStr)) {
if (subStrMap.keySet().contains(subStr)) {
subStrMap.put(subStr, subStrMap.get(subStr) + 1);
} else {
subStrMap.put(subStr, 1);
}
count ++;
if (subStrMap.get(subStr) <= wordMap.get(subStr)) {
count ++;
} else {
// 说明当前开始位置不匹配
while (subStrMap.get(subStr) > wordMap.get(subStr)) {
//
String startWord = S.substring(left, left + wordLen);
subStrMap.put(startWord, subStrMap.get(startWord) - 1);
left += wordLen;
count --;
}
}
if (count == strArr.length) {
// 找到了
result.add(left);
// 向后移动一个单词
count --;
String startWord = S.substring(left, left + wordLen);
subStrMap.put(startWord, subStrMap.get(startWord) - 1);
left += wordLen;
}
} else {
// 清空变量，重新开始查找
left = j + wordLen;
subStrMap.clear();
count = 0;
}
}
}
System.out.println("loopCount------->" + loopCount);
int[] res = new int[result.size()];
for (int i = 0; i < result.size(); i++) {
res[i] = result.get(i);
}
return res;
}
public static void main(String[] args) {
SubstringWithConcatenationOfAllWords substringWithConcatenationOfAllWords = new SubstringWithConcatenationOfAllWords();
String S = "barfoothefoobarman";
String[] strArr = new String[]{"foo", "bar"};
System.out.println(Arrays.toString(substringWithConcatenationOfAllWords.findSubstring(S, strArr)));
System.out.println(Arrays.toString(substringWithConcatenationOfAllWords.findSubstring1(S, strArr)));
}
}
