[LeetCode] Word Break II

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"].

这题跟Word Break I一样，简单的使用递归会超时。结合前一题的解法做剪枝，只对能够被拆解的子字符串进行后续DFS，这样可以有效减少DFS的搜索空间。

可以直接照搬前一版本的代码。

    public List<String> wordBreak(String s, Set<String> dict) {        List<String> ret = new ArrayList<String>(); // An important pruning here, by resuing the version I.// Return an empty collection if no solution exists.if (!isBreakable(s, dict)) {    return ret;}dfs(s, dict, 0, "", ret);return ret;    }        private boolean isBreakable(String s, Set<String> dict) {// dp[i][j] represents the substring that starts at i and ends at j,// where j is exclusive.// dp[*][0] is wasted.boolean[][] dp = new boolean[s.length()][s.length() + 1];for (int len = 1; len <= s.length(); ++len) {for (int start = 0; start <= s.length() - len; ++start) {int end = start + len;String substr = s.substring(start, end);if (len == 1) {dp[start][end] = dict.contains(substr);} else {if (dict.contains(substr)) {dp[start][end] = true;continue;}// Try to partition the word.for (int leftLen = 1; leftLen < len; ++leftLen) {int leftEnd = start + leftLen;if (dp[start][leftEnd] && dp[leftEnd][end]) {dp[start][end] = true;break;}}}}}return dp[0][s.length()];}        private void dfs(String s, Set<String> dict, int start, String sentence, List<String> sentences) {        if (start == s.length()) {            // Remove the leading space before adding it to the result collection.            sentences.add(sentence.substring(1));            return;        }                for (int len = 1; start + len <= s.length(); ++len) {            int end = start + len;            String str = s.substring(start, end);            if (dict.contains(str)) {                dfs(s, dict, end, sentence + " " + str, sentences);            }        }     }