[LeetCode] 140. Word Break II

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思路:
这题做出来以后成就感爆棚啊…第一个是我写的一个标准DFS, 但过不去大样本, 会超时. 第二个版本是借助Word Break http://blog.csdn.net/hiimdaosui/article/details/52457700 里面的DP数组做剪枝, 这样就可以过大样本了.

void dfs(vector<string>& res, string& candidate, string& s, unordered_set<string>& wordDict, int start) {        if (start == s.length()) {            candidate.pop_back();            res.push_back(candidate);            return;        }        for (int i = start; i < s.length(); i++) {            string temp = candidate;            string cur = s.substr(start, i - start + 1);            if (wordDict.count(cur)) {                candidate += cur + " ";                dfs(res, candidate, s, wordDict, i + 1);                candidate = temp;            }        }    }    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {        vector<string> res;        string candidate = "";        dfs(res, candidate, s, wordDict, 0);        return res;    }

void dfs(vector<string>& res, string& candidate, string& s, unordered_set<string>& wordDict, int start, bool breakable[]) {    if (start == s.length()) {        candidate.pop_back();        res.push_back(candidate);        return;    }    for (int i = start; i < s.length(); i++) {        string cur = s.substr(start, i - start + 1);        if (breakable[i + 1] && wordDict.count(cur)) {            string temp = candidate;            candidate += cur + " ";            dfs(res, candidate, s, wordDict, i + 1, breakable);            candidate = temp;        }    }}vector<string> wordBreak(string s, unordered_set<string>& wordDict) {    int slen = s.length();    vector<string> res;    if (! slen) return res;    bool breakable[slen + 1] = {0};    breakable[0] = true;    for (int i = 1; i <= slen; i++) {        for (int j = 0; j < i; j++) {            string cur = s.substr(j, i - j);            if (breakable[j] && wordDict.count(cur)) {                breakable[i] = true;                break;            }        }    }    string candidate = "";    if (breakable[slen])        dfs(res, candidate, s, wordDict, 0, breakable);    return res;}

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