LeetCode-Word Break
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题目:https://oj.leetcode.com/problems/word-break/
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"
.
分析:
动态规划。对字符串s,定义长度为s.length()+1的Boolean型数组(长度为n 的字符串有n+1 个隔板),f[i]表示s[0-i]是否可以分词,则状态转移方程为:f[i]=f[j]&&dict.contains(s[j,i]) , 0<=j<i, 1<i<=s.length
源码:Java版本
算法分析:动态规划。时间复杂度O(n^2),空间复杂度O(n)
public class Solution { public boolean wordBreak(String s, Set<String> dict) { boolean[] f=new boolean[s.length()+1]; f[0]=true; for(int i=1;i<s.length()+1;i++) { for(int j=i-1;j>=0;j--) { if(f[j]&& dict.contains(s.substring(j,i))) { f[i]=true; break; } } } return f[s.length()]; }}
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