|
22 | 22 | > dp[0]=true; |
23 | 23 | >for (int i=1; i<= s.length(); i++) { |
24 | 24 | >for (int j=0; j< i; j++) { |
25 | | -> dp[i]= dp[j]&&wordDict.contains(s.substring(j, i)); |
| 25 | +> dp[i]= dp[j]&&set.contains(s.substring(j, i)); |
26 | 26 | >if (dp[i]) { |
27 | 27 | >break; |
28 | 28 | > } |
@@ -60,7 +60,7 @@ public List<String> wordBreak(String s, List<String> wordDict) { |
60 | 60 | for (int i=1; i<= s.length(); i++) { |
61 | 61 | temp=newArrayList<>(); |
62 | 62 | for (int j=0; j< i; j++) { |
63 | | -if (wordDict.contains(s.substring(j, i))) { |
| 63 | +if (set.contains(s.substring(j, i))) { |
64 | 64 | //得到前半部分的所有情况然后和当前单词相加 |
65 | 65 | for (int k=0; k< dp.get(j).size(); k++) { |
66 | 66 | String t= dp.get(j).get(k); |
@@ -116,7 +116,7 @@ public List<String> wordBreak(String s, List<String> wordDict) { |
116 | 116 | for (int i=1; i<= s.length(); i++) { |
117 | 117 | temp=newArrayList<>(); |
118 | 118 | for (int j=0; j< i; j++) { |
119 | | -if (wordDict.contains(s.substring(j, i))) { |
| 119 | +if (set.contains(s.substring(j, i))) { |
120 | 120 | for (int k=0; k< dp.get(j).size(); k++) { |
121 | 121 | String t= dp.get(j).get(k); |
122 | 122 | if (t.equals("")) { |
|