[LeetCode] 030. Substring with Concatenation of All Words (H
发布时间：2020-12-13 20:15:47 所属栏目：PHP教程 来源：网络整理
导读：索引：[LeetCode] Leetcode 题解索引 (C/Java/Python/Sql) Github: https://github.com/illuz/leetcode 030. Substring with Concatenation of All Words (Hard) 链接 ： 题目：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words
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索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) 030. Substring with Concatenation of All Words (Hard)链接：题目：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/ 题意：给1个字符串 S 和1个单词列表，单词长度都1样，找出所有 S 的子串，子串由所有单词组成，返回子串的起始位置。 分析：很明显每一个子串都是由所有单词组成的，长度是1定的，所以直接枚举子串，切分后再用 map 进行判断就好了。 这样的算法复杂度是 O(n*m)，其实还有几种很酷的 O(n) 的算法：
这里用 C++ 实现了 O(n*m) 算法，用 Java 实现了 1 算法。 代码：C++: class Solution {
public:
vector<int> findSubstring(string S,vector<string> &L) {
map<string,int> words;
map<string,int> curWords;
vector<int> ret;
int slen = S.length();
if (!slen || L.empty()) return ret;
int llen = L.size(),wlen = L[0].length();
// record the current words map
for (auto &i : L)
++words[i];
// check the [llen * wlen] substring
for (int i = 0; i + llen * wlen <= slen; i++) {
curWords.clear();
int j = 0;
// check the words
for (j = 0; j < llen; j++) {
string tmp = S.substr(i + j * wlen,wlen);
if (words.find(tmp) == words.end())
break;
++curWords[tmp];
if (curWords[tmp] > words[tmp])
break;
}
if (j == llen)
ret.push_back(i);
}
return ret;
}
};
Java: public class Solution {
public List<Integer> findSubstring(String S,String[] L) {
List<Integer> ret = new ArrayList<Integer>();
int slen = S.length(),llen = L.length;
if (slen <= 0 || llen <= 0)
return ret;
int wlen = L[0].length();
// get the words' map
HashMap<String,Integer> words = new HashMap<String,Integer>();
for (String str : L) {
if (words.containsKey(str)) {
words.put(str,words.get(str) + 1);
} else {
words.put(str,1);
}
}
for (int i = 0; i < wlen; ++i) {
int left = i,count = 0;
HashMap<String,Integer> tmap = new HashMap<String,Integer>();
for (int j = i; j <= slen - wlen; j += wlen) {
String str = S.substring(j,j + wlen);
if (words.containsKey(str)) {
if (tmap.containsKey(str)) {
tmap.put(str,tmap.get(str) + 1);
} else {
tmap.put(str,1);
}
if (tmap.get(str) <= words.get(str)) {
count++;
} else {
// too many words,push the 'left' forward
while (tmap.get(str) > words.get(str)) {
String tmps = S.substring(left,left + wlen);
tmap.put(tmps,tmap.get(tmps) - 1);
if (tmap.get(tmps) < words.get(tmps)) {
// if affect the count
count--;
}
left += wlen;
}
}
// get the answer
if (count == llen) {
ret.add(left);
// it's better to push forward once
String tmps = S.substring(left,left + wlen);
tmap.put(tmps,tmap.get(tmps) - 1);
count--;
left += wlen;
}
} else {
// not any match word
tmap.clear();
count = 0;
left = j + wlen;
}
}
}
return ret;
}
}
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