标签：style blog color io for cti
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".
题解：动态规划
设置数组dp，dp[i,j]表示从s的第i个字母（i=0~s.length-1）开始，长度为j的子字符串是否在字典dict中。则，有以下递推式：
代码如下：
public class Solution { public boolean wordBreak(String s, Set<String> dict) { if(s == null || dict.size() == 0) return false; int length = s.length(); boolean[][] dp = new boolean[length][length+1]; for(int len = 1;len <= length;len++){ for(int i = 0;i+len <= length;i++){ String sub = s.substring(i,i+len); if(dict.contains(sub)){ dp[i][len] = true; continue; } for(int k = 1;k < len;k++){ if(dp[i][k] && dp[i+k][len-k] ) { dp[i][len] = true; break; } } } } return dp[0][length]; } }
对于字符串，substring这个函数返回的是beginIndex和endIndex-1之间的子字符串！
【leetcode】Word Break,布布扣,bubuko.com
标签：style blog color io for cti
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3849154.html
