題目
https://leetcode.com/problems/word-break-ii/
題解
由 leetcode 139. Word Break | 139. 單詞拆分（動態規劃） 改造而來。
dp 過程示例:
class Solution {public List
<String> wordBreak(String s
, List
<String> wordDict
) {HashSet
<String> dict
= new HashSet<>();for (String word
: wordDict
) {dict
.add(word
);}int n
= s
.length();ArrayList
<int[]>[][] dp
= new ArrayList[n
+ 1][n
+ 1];for (int i
= 1; i
<= n
; i
++) {for (int j
= i
; j
<= n
; j
++) {int L
= j
- i
;int R
= j
;dp
[L
][R
] = new ArrayList<>();if (dict
.contains(s
.substring(L
, R
))) {ArrayList
<int[]> t
= new ArrayList<>();t
.add(new int[]{L
, R
});dp
[L
][R
] = t
;}for (int M
= L
+ 1; M
< R
; M
++) {if (dp
[L
][M
] != null
&& dp
[L
][M
].size() > 0 && dp
[M
][R
] != null
&& dp
[M
][R
].size() > 0) {dp
[L
][R
].add(new int[]{L
, M
});dp
[L
][R
].add(new int[]{M
, R
});}}}}Set
<String> res
= join(dp
, 0, n
, s
);return new ArrayList<>(res
);}public Set
<String> join(ArrayList
<int[]>[][] dp
, int L
, int R
, String s
) {ArrayList
<int[]> pair
= dp
[L
][R
];Set
<String> res
= new HashSet<>();int i
= 0;if (pair
.size() % 2 == 1) {res
.add(s
.substring(L
, R
));i
= 1;}for (; i
+ 1 < pair
.size(); i
+= 2) {Set
<String> lSet
= join(dp
, L
, pair
.get(i
)[1], s
);Set
<String> rSet
= join(dp
, pair
.get(i
+ 1)[0], R
, s
);for (String l
: lSet
) {for (String r
: rSet
) {res
.add(l
+ " " + r
);}}}return res
;}
}
總結
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