Word Break

Med Given a string s and a dictionary of words dict, determine if s can be break into a space-separated sequence of one or more dictionary words.

Thoughts

问给定的str能否由字典内元素拼凑而成。也就是找是否存在划分，action为枚举可能的划分，DP。

Code

/*
 * @lc app=leetcode id=139 lang=cpp
 *
 * [139] Word Break
 */

// @lc code=start
class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> words(wordDict.begin(), wordDict.end());
        const int N = s.size();
        int mx_len = 0;
        for (const auto w : wordDict) {
            mx_len = max((int)w.length(), mx_len);
        }
        vector<bool> dp(N + 1, false);
        dp[0] = true;
        for (int i = 1; i <= N; ++i) {
            for (int j = max(i - 1 - mx_len, 0); j < i; ++j) {
                if (dp[j] && words.count(s.substr(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[N];
    }
};
// @lc code=end

public class Solution {
    /**
     * @param s: A string s
     * @param dict: A dictionary of words dict
     */
    public boolean wordBreak(String s, Set<String> dict) {
        int n = s.length();
        // init 
        boolean[] f = new boolean[n + 1];
        f[0] = true; // empty, 前0个
        int maxLen = 0;
        for (String str : dict) {
            maxLen = Math.max(maxLen, str.length());
        }

        // loop
        for (int i = 1; i <= n; i++) {
            for (int j = Math.max(0, i - maxLen); j < i; j++) {
                if (f[j] && dict.contains(s.substring(j, i))) {
                    f[i] = true;
                    break;
                }
            }
        }

        return f[n];
    }
}

Analysis

TC: $O(n^2)$

需要作一点小优化，否则l**tcode会超时。