LeetCode.Word Break

試題請參見: https://leetcode.com/problems/word-break/

題目概述

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".

解題思路

DP問題.

使用數組isMatched[i]表示字符串s[0, i - 1]是否可以被劃分.

因此isMatched[s.length]表示字符串s是否可以被劃分.

若isMatched[i] == true, 表示s[0, i - 1]可以被劃分, 則需要同時滿足如下條件之一:

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

5. 對於isMatched[5], 因爲"aaaaa"不在dict中

   • 進而需要判斷isMatched[1] = true 且 s[1...4]在dict中, 顯然前者不成立
   • 進而需要判斷isMatched[2] = true 且 s[2...4]在dict中, 顯然前者不成立
   • 進而需要判斷isMatched[3] = true 且 s[3...4]在dict中, 顯然後者不成立
   • 進而需要判斷isMatched[4] = true 且 s[4...4]在dict中, 顯然後者不成立

因此isMatched[5] == false.

源代碼

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));
    }
}