LeetCode.Word Break

试题请参见: https://leetcode.com/problems/word-break/

题目概述

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".

解题思路

DP问题.

使用数组isMatched[i]表示字符串s[0, i - 1]是否可以被划分.

因此isMatched[s.length]表示字符串s是否可以被划分.

若isMatched[i] == true, 表示s[0, i - 1]可以被划分, 则需要同时满足如下条件之一:

• isMatched[0] == true, 且s[0, i - 1]在dict中
• isMatched[1] == true, 且s[1, i - 1]在dict中
• isMatched[2] == true, 且s[2, i - 1]在dict中
• `…
• isMatched[i - 1] == true, 且s[i - 1, i - 1]在dict中

例如: s = "aaaaa", dict = { "aaaa", "aaa" }.

1. isMatched[1] == false, 因为"a"不在dict中
2. isMatched[2] == false, 因为"aa"不在dict中
3. isMatched[3] == true, 因为"aaa"在dict中
4. isMatched[4] == true, 因为"aaaa"在dict中

5. 对于isMatched[5], 因为"aaaaa"不在dict中

   • 进而需要判断isMatched[1] = true 且 s[1...4]在dict中, 显然前者不成立
   • 进而需要判断isMatched[2] = true 且 s[2...4]在dict中, 显然前者不成立
   • 进而需要判断isMatched[3] = true 且 s[3...4]在dict中, 显然后者不成立
   • 进而需要判断isMatched[4] = true 且 s[4...4]在dict中, 显然后者不成立

因此isMatched[5] == false.

源代码

import java.util.HashSet;import java.util.Set;public class Solution {    public boolean wordBreak(String s, Set<String> wordDict) {        int length = s.length();        boolean[] isMatched = new boolean[length + 1];        isMatched[0] = true;        for ( int i = 1; i  <= length; ++ i ) {            for ( int j = 0; j < i; ++ j ) {                if ( isMatched[j] && wordDict.contains(s.substring(j, i)) ) {                    isMatched[i] = true;                    break;                }            }        }        return isMatched[length];    }    public static void main(String[] args) {        Solution s = new Solution();        Set<String> set = new HashSet<String>();        set.add("leet");        set.add("code");        System.out.println(s.wordBreak("leetcode", set));    }}