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" }

.

isMatched[1] == false

, 因为

"a"

不在dict中

isMatched[2] == false

, 因为

"aa"

不在dict中

isMatched[3] == true

, 因为

"aaa"

在dict中

isMatched[4] == true

, 因为

"aaaa"

在dict中

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));
}
}