LeetCode: Substring with Concatenation of All Words 解题报告

[b]Substring with Concatenation of All Words[/b]

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).

// SOLUTION 2:
public List<Integer> findSubstring(String S, String[] L) {
HashMap<String, Integer> map = new HashMap<String, Integer>();
HashMap<String, Integer> found;
List<Integer> ret = new ArrayList<Integer>();

if (S == null || L == null || L.length == 0) {
return ret;
}

// put all the strings into the map.
for (String s: L) {
if (map.containsKey(s)) {
map.put(s, map.get(s) + 1);
} else {
map.put(s, 1);
}
}

int lenL = L[0].length();

// 注意这里的条件：i < S.length() - lenL * L.length
// 这里很关键，如果长度不够了，不需要再继续查找
for (int i = 0; i <= S.length() - lenL * L.length; i++) {
// 每一次，都复制之前的hashMap.
found = new HashMap<String, Integer>(map);

// 一次前进一个L的length.
// 注意j <= S.length() - lenL; 防止越界
for (int j = i; j <= S.length() - lenL; j += lenL) {
String sub = S.substring(j, j + lenL);
if (found.containsKey(sub)) {
// 将找到字符串的计数器减1.
found.put(sub, found.get(sub) - 1);

// 减到0即可将其移出。否则会产生重复运算，以及我们用MAP为空来判断是否找到所有的单词。
if (found.get(sub) == 0) {
found.remove(sub);
}
} else {
// 不符合条件，可以break,i前进到下一个匹配位置
break;
}

// L中所有的字符串都已经找到了。
if (found.isEmpty()) {
ret.add(i);
}
}
}

return ret;
}

View Code
[b]SOLUTION 3:[/b]

九章算法官网解：

http://www.ninechapter.com/solutions/substring-with-concatenation-of-all-words/

[b]主页君GITHUB:[/b]

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/FindSubstring.java