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"].
[解题思路]
DFS + backtracking
在大数据："aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]
挂掉了
1 public static ArrayList<String> wordBreak(String s, Set<String> dict) { 2 // Note: The Solution object is instantiated only once and is reused by 3 // each test case. 4 ArrayList<String> result = new ArrayList<String>(); 5 if (s == null || s.length() == 0) { 6 return result; 7 } 8 int length = s.length(); 9 dfs(s, 0, 0, length, dict, result, new StringBuffer()); 10 return result; 11 } 12 13 public static void dfs(String s, int start, int depth, int length, 14 Set<String> dict, ArrayList<String> result, StringBuffer sb) { 15 if (depth == length) { 16 String t = sb.toString(); 17 result.add(t.substring(0, t.length() - 1)); 18 return; 19 } 20 21 for (int len = 1; len <= length - start; len++) { 22 String t = s.substring(start, start + len); 23 if (dict.contains(t)) { 24 int beforeAddLen = sb.length(); 25 sb.append(t).append(" "); 26 dfs(s, start + len, depth + len, length, dict, result, sb); 27 sb.delete(beforeAddLen, sb.length()); 28 } 29 } 30 }
2.DP+DFS
使用Word Break对字符串进行处理，处理结束后使用DFS进行DFS
1 public class Solution { 2 public static ArrayList<String> wordBreak(String s, Set<String> dict) { 3 ArrayList<String> result = new ArrayList<String>(); 4 if (s == null || dict.size() <= 0) { 5 return result; 6 } 7 int length = s.length(); 8 // seg(i, j) means substring t start from i and length is j can be 9 // segmented into 10 // dictionary words 11 boolean[][] seg = new boolean[length][length + 1]; 12 for (int len = 1; len <= length; len++) { 13 for (int i = 0; i < length - len + 1; i++) { 14 String t = s.substring(i, i + len); 15 if (dict.contains(t)) { 16 seg[i][len] = true; 17 continue; 18 } 19 for (int k = 1; k < len; k++) { 20 if (seg[i][k] && seg[i + k][len - k]) { 21 seg[i][len] = true; 22 break; 23 } 24 } 25 } 26 } 27 if (!seg[0][length]) { 28 return result; 29 } 30 31 int depth = 0; 32 dfs(s, seg, 0, length, depth, result, new StringBuffer(), dict); 33 34 return result; 35 } 36 37 private static void dfs(String s, boolean[][] seg, int start, int length, 38 int depth, ArrayList<String> result, StringBuffer sb, Set<String> dict) { 39 if (depth == length) { 40 String t = sb.toString(); 41 result.add(t.substring(0, t.length() - 1)); 42 return; 43 } 44 45 for (int len = 1; len <= length; len++) { 46 if (seg[start][len]) { 47 String t = s.substring(start, start + len); 48 if(!dict.contains(t)){ 49 continue; 50 } 51 int beforeAddLen = sb.length(); 52 sb.append(t).append(" "); 53 dfs(s, seg, start + len, length, start + len, result, sb, dict); 54 sb.delete(beforeAddLen, sb.length()); 55 } 56 } 57 } 58 }