30.Substring with Concatenation of All Words

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].

(order does not matter).

public List<integer> findSubstring(String s, String[] words) {
	int N = s.length();
	List<integer> indexes = new ArrayList<integer>(s.length());
	if (words.length == 0) {
		return indexes;
	}
	int M = words[0].length();
	if (N < M * words.length) {
		return indexes;
	}
	int last = N - M + 1;
	// 很简单的数据处理
	
	//map each string in words array to some index and compute target counters
	Map<string integer=""> mapping = new HashMap<string integer="">(words.length);
	int [][] table = new int[2][words.length];
	int failures = 0, index = 0;
	for (int i = 0; i < words.length; ++i) {
		Integer mapped = mapping.get(words[i]);
		if (mapped == null) {
			++failures;
			mapping.put(words[i], index);
			mapped = index++;
		}
		++table[0][mapped];
	}
	// 将题目中的数据放到目标容器中
	
	//find all occurrences at string S and map them to their current integer, -1 means no such string is in words array
	int [] smapping = new int[last];
	for (int i = 0; i < last; ++i) {
		String section = s.substring(i, i + M);
		Integer mapped = mapping.get(section);
		if (mapped == null) {
			smapping[i] = -1;
		} else {
			smapping[i] = mapped;
		}
	}
	// 到这里题目就变成对数组的处理了，真的不得不说原作者的思路真的很巧
	
	//fix the number of linear scans
	for (int i = 0; i < M; ++i) {
		//reset scan variables
		int currentFailures = failures; //number of current mismatches
		int left = i, right = i;
		Arrays.fill(table[1], 0);
		//here, simple solve the minimum-window-substring problem
		while (right < last) {
			while (currentFailures > 0 && right < last) {
				int target = smapping[right];
				if (target != -1 && ++table[1][target] == table[0][target]) {
					--currentFailures;
				}
				right += M;
			}
			while (currentFailures == 0 && left < right) {
				int target = smapping[left];
				if (target != -1 && --table[1][target] == table[0][target] - 1) {
					int length = right - left;
					//instead of checking every window, we know exactly the length we want
					if ((length / M) ==  words.length) {
						indexes.add(left);
					}
					++currentFailures;
				}
				left += M;
			}
		}
		
	}
	return indexes;
}
// 如果对过程不理解的，可以去实际测试数据并调试或中间输出。
// 耽误时间好久~</string></string></integer></integer></integer>