Created
March 23, 2014 19:53
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Word Break @ LeetCode
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public class Solution { | |
public boolean wordBreak(String s, Set<String> dict) { | |
if(s == null || s.length() == 0 || dict == null) | |
return true; | |
int length = s.length(); | |
//dp[i] is true when the substring from i to the end can be partitioned | |
boolean[] dp = new boolean[length+1]; | |
for(boolean b : dp) | |
b = false; | |
//empty string can be partitioned for sure | |
dp[length] = true; | |
//bottom up dp, start from the end | |
for(int i = length - 1;i>=0;i--){ | |
for(int j = i;j < length;j++){ | |
String sub = s.substring(i,j+1); | |
if(dict.contains(sub) == true && dp[j+1] == true){ | |
dp[i]= true; | |
break; //break, just jump out current level of loop | |
} | |
} | |
} | |
return dp[0]; | |
} | |
} | |
/*********recursive solution***********/ | |
public class Solution { | |
public boolean wordBreak(String s, Set<String> dict) { | |
if(dict.contains(s) == true) | |
return true; | |
int length = s.length(); | |
for(int i = 1;i < length;i++){ | |
String firstPart = s.substring(0,i); | |
if(dict.contains(firstPart) == true){ | |
String remainPart = s.substring(i,length); | |
boolean result = wordBreak(remainPart, dict); | |
if(result == true){ | |
return true; | |
} | |
} | |
} | |
return false; | |
} | |
} |
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