字歇(用动态规划:上>下)给定一个字符串s和文字的字典dict,以s加空格来构造一个句子 其中每个字是一个有效的字典中的单词。
返回所有这些可能的句子。
例如,因为 S =catsanddog,字典=猫,猫,和,沙,狗。
一个解决方案是[猫与狗,猫沙狗。 问:
在时间复杂度? 在空间的复杂性?我个人认为,
在时间复杂度= O(N!),没有动态规划,n是给定字符串的长度, 空间复杂度= O(N)。的疑惑:
想不通与动态规划的时间复杂度。 在它看来,上方的空间复杂度是不正确的。code 【JAVA]
公共类解决方案{
公开名单<字符串> wordBreak(字符串S,集<字符串>字典){
名单<字符串>名单=新的ArrayList<字符串>();
//输入检查。
如果(S == NULL || s.length()== 0 ||
字典== NULL || dict.size()== 0)返回目录;
INT的len = s.length();
//备忘录[i]的记录,
//我们是否切断指数我,就可以得到结果之一。
布尔备忘录[] =新的布尔[LEN]
的for(int i = 0; I< LEN;我++)备注[I] = TRUE;
StringBuilder的tmpStrBuilder =新的StringBuilder();
助手(S,0,tmpStrBuilder,字典,列表,备忘录);
返回列表;
}
私人无效帮手(的String,诠释开始,StringBuilder的tmpStrBuilder,
设置<字符串>字典,列表和LT;字符串>列表中,布尔[]备忘录){
//基本情况。
如果(开始> = s.length()){
list.add(tmpStrBuilder.toString()修剪());
返回;
}
INT listSizeBeforeRecursion = 0;
的for(int i =启动; I< s.length();我++){
如果(备注[I] ==假)继续;
字符串CURR = s.substring(启动,I + 1);
如果继续(dict.contains(CURR)!);
//有一个尝试。
tmpStrBuilder.append(CURR);
tmpStrBuilder.append();
//做递归。
listSizeBeforeRecursion =则为list.size();
助手(S,I + 1,tmpStrBuilder,字典,列表,备忘录);
如果(则为list.size()== listSizeBeforeRecursion)备注[I] = FALSE;
//回滚。
tmpStrBuilder.setLength(tmpStrBuilder.length() - curr.length() - 1);
}
}
}
解决方案
通过DP:
时间:O(N * M) N - 字符串大小 米 - 字典大小
内存:O(N)
请参阅我的答案在这里,用code例如:
Dynamic编程 - 字歇
Word Break(with Dynamic Programming: Top->Down) 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"]. Question:
Time complexity ? Space complexity ?Personally I think,
Time complexity = O(n!), without Dynamic Programming, n is the length of the given string, Space complexity = O(n).The puzzled:
Can not figure out the time complexity with Dynamic Programming. It seems that the space complexity above is not correct.Code[Java]
public class Solution {
public List<String> wordBreak(String s, Set<String> dict) {
List<String> list = new ArrayList<String>();
// Input checking.
if (s == null || s.length() == 0 ||
dict == null || dict.size() == 0) return list;
int len = s.length();
// memo[i] is recording,
// whether we cut at index "i", can get one of the result.
boolean memo[] = new boolean[len];
for (int i = 0; i < len; i ++) memo[i] = true;
StringBuilder tmpStrBuilder = new StringBuilder();
helper(s, 0, tmpStrBuilder, dict, list, memo);
return list;
}
private void helper(String s, int start, StringBuilder tmpStrBuilder,
Set<String> dict, List<String> list, boolean[] memo) {
// Base case.
if (start >= s.length()) {
list.add(tmpStrBuilder.toString().trim());
return;
}
int listSizeBeforeRecursion = 0;
for (int i = start; i < s.length(); i ++) {
if (memo[i] == false) continue;
String curr = s.substring(start, i + 1);
if (!dict.contains(curr)) continue;
// Have a try.
tmpStrBuilder.append(curr);
tmpStrBuilder.append(" ");
// Do recursion.
listSizeBeforeRecursion = list.size();
helper(s, i + 1, tmpStrBuilder, dict, list, memo);
if (list.size() == listSizeBeforeRecursion) memo[i] = false;
// Roll back.
tmpStrBuilder.setLength(tmpStrBuilder.length() - curr.length() - 1);
}
}
}
解决方案
With DP:
Time: O(N*M) N - string size M - dict size
Memory: O(N)
See my answer here, with code example:
Dynamic Programming - Word Break