[leetcode 140] Word Break II

最新推荐文章于 2025-12-31 20:36:24 发布

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了如何利用深度学习算法实现高效准确的视频分析。

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        const int n = s.size();
        int f[n+1];
        fill_n(&f[0], n+1, false);
        f[0] = true;
        vector<vector<bool> > prev(n+1, vector<bool>(n, false));
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (f[j] && dict.find(s.substr(j,i-j)) != dict.end()) {
                    f[i] = true;
                    prev[i][j] = true;
                }
            }
        }
        vector<string> cur;
        vector<string> res;
        dfs(s, n, cur, res, prev);
        return res;
        
    }
    void dfs(string s, int n, vector<string> &cur, vector<string> &res, const vector<vector<bool> > &prev) {
        if (n == 0) {
            string tmp;
            for (auto i = cur.crbegin(); i != cur.crend(); i++) {
                tmp += *i + " ";
            }
            tmp.erase(tmp.end()-1);
            res.push_back(tmp);
            return ;
        }
        for (int i = s.size()-1; i >= 0; i--) {
            if (prev[n][i]) {
                cur.push_back(s.substr(i, n-i));
                dfs(s, i, cur, res, prev);
                cur.pop_back();
            }
        }
    }
};