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Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "leetcode"
,
dict = ["leet", "code"]
.
Return true because "leetcode"
can be segmented as "leet
code"
.
public boolean wordBreak(String s, Set<String> dict) { StringBuilder sb = new StringBuilder(); return helper(s, dict, 0, sb); } private boolean helper(String s, Set<String> dict, int index, StringBuilder sb) { if (index == dict.size()) { return s.equals(sb.toString()); } Iterator<String> iter = dict.iterator(); while (iter.hasNext()) { StringBuilder sb1 = new StringBuilder(sb); String dic = iter.next(); sb.append(dic); iter.remove(); if (helper(s, dict, index+1, sb)) return true; dict.add(dic); sb = sb1; } return false; }
public boolean wordBreak(String s, Set<String> dict) { StringBuilder sb = new StringBuilder(); return helper(s, dict, 0, sb); } private boolean helper(String s, Set<String> dict, int index, StringBuilder sb) { if (!s.startsWith(sb.toString())) { return false; } if (index == dict.size()) { return s.equals(sb.toString()); } Iterator<String> iter = dict.iterator(); while (iter.hasNext()) { StringBuilder sb1 = new StringBuilder(sb); String dic = iter.next(); sb.append(dic); iter.remove(); if (helper(s, dict, index+1, sb)) return true; dict.add(dic); sb = sb1; } return false; }其实这个剪枝可以减去很多分支,但是还是超时!怎么办?只能用最高效的动态规划了,上代码:
public boolean wordBreak1(String s, Set<String> dict) { int length = s.length(); boolean[] can = new boolean[length+1]; can[0] = true; for (int i = 1; i <= length; i++) { for (int j = 0; j < i; j++) { if (can[j] && dict.contains(s.substring(j, i))) { can[i] = true; break; } } } return can[length]; }boolean数组can[i]是指,s.substring(0,i)是可以分割的,所示数组最后一个元素即为所求。
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原文地址:http://blog.csdn.net/my_jobs/article/details/44245765