leetcode 139. Word Break 动态规划

题意

输入：(String s, List wordDict)
输出：s是否可有wordDict中的单词拼凑而成。可以重复使用。

思路

动态规划，一维数组存前n位所组成的字符串的结果。
dp[n] = 任意一个i < n，有dp[i] && 右边的字符串在wordDict中出现。

解1

    // Runtime: 6 ms, faster than 64.73% of Java online submissions for Word Break.
    //Memory Usage: 39.7 MB, less than 10.64% of Java online submissions for Word Break.
    public boolean wordBreak(String s, List wordDict) {
        Set dict = new HashSet<>(wordDict);

        boolean[] rst = new boolean[s.length() + 1];
        rst[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int k = 0; k < i; k++) {
                if (rst[k] && dict.contains(s.substring(k, i))) {
                    rst[i] = true;
                    break;
                }
            }
        }
        return rst[s.length()];
    }

解2

思路和解1一样，换一种字典搜索方式，效率提高一些。

    // Runtime: 3 ms, faster than 84.19% of Java online submissions for Word Break.
    //Memory Usage: 37.2 MB, less than 10.64% of Java online submissions for Word Break.
    public boolean wordBreak(String s, List wordDict) {
        Map> sizedWordDict = new HashMap<>();
        for (String word : wordDict) {
            List words = sizedWordDict.computeIfAbsent(word.length(), k -> new ArrayList<>());
            words.add(word);
        }

        boolean[] rst = new boolean[s.length() + 1];
        rst[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int k = 0; k < i; k++) {
                if (rst[k]) {
                    List words = sizedWordDict.get(i - k);
                    if (words != null && words.contains(s.substring(k, i))) {
                        rst[i] = true;
                        break;
                    }
                }
            }
        }
        return rst[s.length()];
    }