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"].
[分析] 最能感受dp剪枝好处的例子：
s="aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab",
dict = ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa",
"aaaaaaaaa","aaaaaaaaaa"]
//////////////////method 3////////////////////
//tag: dp + dfs/backtracking/recursion
//dp[i] means s.substring(i, N) satisfy wordBreak
//dp[i] |= s.substring(i, j) && dp[j], i < j <=N
public List<String> wordBreak(String s, Set<String> dict) {
ArrayList<String> result = new ArrayList<String>();
if (dict == null || dict.size() == 0 || s == null || s.length() == 0)
return result;
int N = s.length();
boolean[] dp = new boolean[N + 1];
dp[N] = true;
for (int i = N - 1; i >= 0; i--) {
StringBuilder sub = new StringBuilder(s.substring(i, N));
for (int j = N; j >= i; j--) {
//dp[i]: s.sub(i, N), sub: s.sub(i, j), dp[j]: s.sub(j, N)
dp[i] = dp[j] && dict.contains(sub.toString());
if (dp[i])
break;
if (j > i)
sub.deleteCharAt(sub.length() - 1);
}
}
if (dp[0]) {
StringBuilder item = new StringBuilder();
recur(s, dict, 0, dp, item, result);
}
return result;
}
// 使用dp[]进行剪枝, dfs/backtracking/recursion
// item = s.sub(0, beg)
public void recur(String s, Set<String> dict, int beg, boolean[] dp, StringBuilder item, ArrayList<String> result) {
if (beg == s.length()) {
item.deleteCharAt(item.length() - 1);
result.add(item.toString());
return;
}
for (int i = beg + 1; i <= s.length(); i++) {
String sub = s.substring(beg, i);
if(dict.contains(sub)&& dp[i]) {
int oldLength = item.length();
item.append(sub).append(' ');
recur(s, dict, i, dp, item, result);
item.delete(oldLength, item.length());
}
}
}
/////////////////////method 2////////////////////
/* Time Limit Exceeded
public List<String> wordBreak(String s, Set<String> dict) {
ArrayList<String> result = new ArrayList<String>();
if (s == null || s.length() == 0)
return result;
int N = s.length();
boolean[] dp = new boolean[N + 1];
ArrayList<ArrayList<StringBuilder>> data = new ArrayList<ArrayList<StringBuilder>>(N + 1);
dp[0] = true;
ArrayList<StringBuilder> data0 = new ArrayList<StringBuilder>();
data0.add(new StringBuilder());
data.add(data0);
for (int i = 1; i <= N; i++) {
data.add(new ArrayList<StringBuilder>());
StringBuilder sub = new StringBuilder(s.substring(0, i));
for (int j = 0; j < i; j++) {
if (dp[j] && dict.contains(sub.toString())) {
dp[i] = true;
for (StringBuilder sb : data.get(j)) {
StringBuilder copy = new StringBuilder(sb.toString());
copy.append(sub.toString()).append(' ');
data.get(i).add(copy);
}
}
sub.deleteCharAt(0);
}
}
for (StringBuilder sb : data.get(N)) {
sb.deleteCharAt(sb.length() - 1);
result.add(sb.toString());
}
return result;
}
*/
//////////////////method 1/////////////////////
/* Time Limit Exceeded
public List<String> wordBreak(String s, Set<String> dict) {
ArrayList<String> result = new ArrayList<String>();
if(s == null || s.length() == 0)
return result;
recur(s, dict, 0, new StringBuilder(), result);
return result;
}
// 太多冗余计算
public void recur(String s, Set<String>dict, int start, StringBuilder item, ArrayList<String> result) {
if(start == s.length()) {
item.deleteCharAt(item.length() - 1);
result.add(item.toString());
return;
}
for (int i = start; i < s.length(); i++) {
String sub = s.substring(start, i + 1);
if (dict.contains(sub)) {
int oldLength = item.length();
item.append(sub).append(' ');
recur(s, dict, i + 1, item, result);
int newLength = item.length();
item.delete(oldLength, newLength);
}
}
}
*/
}