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题目地址：https://leetcode.com/problems/word-break-ii/
题目解析：看到题目的第一思路是采用递归暴力解法，每找到一个单词将单词添加到返回的结果集中，并将查找的开始位置后移直到字符串的结尾。
题目解答：
import java.util.HashSet; import java.util.LinkedList; import java.util.List; import java.util.Set; public class Solution { public List<String> wordBreak(String s, Set<String> wordDict) { List<String> res = new LinkedList<String>(); if(s == null || s.length() == 0){ return res; } wordBreak(s, 0, new StringBuilder(), wordDict, res); return res; } private void wordBreak(String s,int start,StringBuilder temp,Set<String> dict,List<String> ret){ if(start>=s.length()){ ret.add(temp.toString()); return; } int len = temp.length(); for(int i=start;i<s.length();i++){ if(dict.contains(s.substring(start, i+1))){ temp.append(temp.length() == 0 ?s.substring(start, i+1):(" "+s.substring(start, i+1))); wordBreak(s, i+1, temp, dict, ret); temp.delete(len, temp.length()); } } } }
这个解法的思路很简单，但是中间有很多重复的步骤，而且这些重复的步骤有有很多字符串操作比较费时，故考虑先使用动态规划找到能break的点，然后以能break点来遍历找到所有break的路径，解答如下：
import java.util.HashSet; import java.util.LinkedList; import java.util.List; import java.util.Set; public class Solution { public static List<String> wordBreak(String s, Set<String> dict) { if (s == null || s.length() == 0 || dict == null) { return null; } List<String> ret = new ArrayList<String>(); List<String> path = new ArrayList<String>(); int len = s.length(); boolean[] D = new boolean[len + 1]; D[len] = true; for(int i=len-1;i>=0;i--){ D[i] = false; for(int j=i;j<len;j++){ if(D[j+1] && dict.contains(s.substring(i,j+1))){ D[i] = true; break; } } } dfs(s, dict, path, ret, 0, D); return ret; } public static void dfs(String s, Set<String> dict, List<String> path, List<String> ret, int index, boolean canBreak[]) { int len = s.length(); if (index == len) { StringBuilder sb = new StringBuilder(); for (String str: path) { sb.append(str); sb.append(" "); } sb.deleteCharAt(sb.length() - 1); ret.add(sb.toString()); return; } if (!canBreak[index]) { return; } for (int i = index; i < len; i++) { String left = s.substring(index, i + 1); if (dict.contains(left)) { path.add(left); dfs(s, dict, path, ret, i + 1, canBreak); path.remove(path.size() - 1); } } } }
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原文地址：http://www.cnblogs.com/xiongyuesen/p/4440497.html
