Substring with Concatenation of All Words -- leetcode

标签：leetcode   面试   子串   拼接   查找   

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).

利用滚动Hash，接合窗口移动，完成下面这个算法。

思想是，先计算出Hash值，用Hash值完成在map中的查找，最后用字符串精确匹配最验证。

并尽可能推迟并省掉字符串精确匹配。

这个算法虽然被leetcode AC了。 但是实际运行时间并预想的慢多了，高达1272ms。

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        vector<int> result;
        if (L.empty()) return result;
        const int itemSize = L[0].size();
        const int wndSize = itemSize * L.size();

        const int q = 3355439;
        const int r = 256;
        map<int, vector<int> > map1;
        for (int i=0; i<L.size(); i++) {
                int nHash = 0;
                for (int j=0; j<L[i].size(); j++) {
                        nHash = ((nHash * r) % q + L[i][j]) % q;
                }
                map1[nHash].push_back(i);
        }

        vector<int> total(itemSize);
        vector<int> revMap(wndSize);
        vector<pair<int,bool> > map2(wndSize);
        int head2 = 0;
        int weight = 1;

        for (int i=1; i<itemSize; i++)
                weight = (weight * r) % q;

        int nHash = 0;
        for (int i=0; i<itemSize-1; i++)
                nHash = ((nHash * r) % q + S[i]) % q;

        for (int i=itemSize-1; i<S.size(); i++) {
                const int group = i % itemSize;
                const int offset = group * L.size();

                if (map2[head2].first) {
                        revMap[offset + map2[head2].first - 1] = 0;
                        map2[head2].first = 0;
                        map2[head2].second = false;
                        --total[group];
                }

                nHash = ((nHash * r) % q + S[i]) % q;
                map<int, vector<int> >::iterator iter = map1.find(nHash);
                if (iter != map1.end()) {
                        int pos = -1;
                        bool compare = true;
                        const vector<int> &val = (*iter).second;
                        if (val.size() == 1) {
                                const int entry = revMap[offset + val[0]];
                                if (!entry ||
                                        !S.compare(i-itemSize+1, itemSize, L[val[0]])) {
                                        pos = 0;
                                        compare = !!entry;
                                }
                        }
                        else {
                                compare = true;
                                int min = i+1;
                                for (int j=0; j<val.size(); j++) {
                                        if (!S.compare(i-itemSize+1, itemSize, L[val[j]])) {
                                                const int entry = revMap[offset + val[j]];
                                                if (!entry) {
                                                        pos = j;
                                                        break;
                                                }
                                                else if (entry < min) {
                                                        min = entry;
                                                        pos = j;
                                                }
                                        }
                                }
                        }
                        
                        if (pos != -1) {
                                const int entry = revMap[offset + val[pos]];

                                revMap[offset + val[pos]] = i + 1;
                                map2[head2].first = val[pos] + 1;
                                map2[head2].second= compare;
                                total[group]++;

                                if (entry) {
                                        const int entry2 = (head2 - i + entry - 1 + wndSize) % wndSize;
                                        map2[entry2].first = 0;
                                        map2[entry2].second= false;
                                        total[group]--;
                                }

                                if (total[group] == L.size()) {
                                        for (int j=0; j<L.size(); j++) {
                                                const int entry2 = (head2 - i + revMap[offset+j] - 1 + wndSize) % wndSize;
                                                if (!map2[entry2].second) {
                                                        if (!S.compare(revMap[offset+j]-itemSize, itemSize, L[j])) {
                                                                map2[entry2].second = true;
                                                        }
                                                        else {
                                                                revMap[offset+j] = 0;
                                                                map2[entry2].first = 0;
                                                                map2[entry2].second= false;
                                                                total[group]--;
                                                                break;
                                                        }
                                                }
                                        }

                                        if (total[group] == L.size())
                                                result.push_back(i-wndSize+1);
                                }
                        }
                }
                
                nHash = ((nHash - (S[i-itemSize+1] * weight) % q) % q + q) % q;
                head2 = (head2 + 1) % wndSize;
        }

        return result;
    }
};

Substring with Concatenation of All Words -- leetcode

标签：leetcode   面试   子串   拼接   查找   

原文地址：http://blog.csdn.net/elton_xiao/article/details/42364383