[LeetCode] 030. Substring with Concatenation of All Words (Hard) (C++/Java)，leetcode

[LeetCode] 030. Substring with Concatenation of All Words (Hard) (C++/Java)，leetcode

索引：[LeetCode] Leetcode 題解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

030. Substring with Concatenation of All Words (Hard) 連結：

題目：https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/
代碼(github)：https://github.com/illuz/leetcode

題意：

給一個字串 S 和一個單字清單，單詞長度都一樣，找出所有 S 的子串，子串由所有單片語成，返回子串的起始位置。

分析：

很明顯每個子串都是由所有單片語成的，長度是一定的，所以直接枚舉子串，切分後再用 map 進行判斷就行了。

這樣的演算法複雜度是 O(n*m)，其實還有幾種很酷的 O(n) 的演算法：

這裡用 C++ 實現了 O(n*m) 演算法，用 Java 實現了 1 演算法。

代碼：

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 mapfor (auto &i : L)++words[i];// check the [llen * wlen] substringfor (int i = 0; i + llen * wlen <= slen; i++) {curWords.clear();int j = 0;// check the wordsfor (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;    }}