lecode Interleaving String

 

这个问题，前面思考过，当时就是用搜索的方法，此处又遇到一次，发现自己理解的太浅了

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example, Given: s1 = "aabcc", s2 = "dbbca",

When s3 = "aadbbcbcac", return true. When s3 = "aadbbbaccc", return false.

1.搜索的方法(超时)

 1 public class Solution {

 2     public boolean isInterleave(String s1, String s2, String s3) {

 3          return isInter(s1,s2,s3,0,0,0);

 4         

 5         

 6     }

 7     public boolean isInter(String s1,String s2,String s3,int r1,int r2, int r3)

 8     {

 9         if(r3==s3.length()) return true;

10         boolean ans=false;

11         

12         if(r1<s1.length()&&s1.charAt(r1)==s3.charAt(r3))

13         {

14            ans=isInter(s1,s2,s3,r1+1,r2,r3+1);

15         

16         

17         

18         }

19         if(ans) return true;

20         

21         if(r2<s2.length()&&s2.charAt(r2)==s3.charAt(r3))

22         {

23            ans=isInter(s1,s2,s3,r1,r2+1,r3+1);

24            return ans;

25         

26         

27         

28         }

29         

30         return false;

31         

32         

33     }

34 }

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dp[i][j]表示s1前i 和s2前j个是否能组成s3的前i+j+1个,  false 不能  true 能

dp[s1.len-1][s2.len-1] 就是我们的答案 

dp[i][j]=dp[i-1][j]&&s1[i]==s3[i+j+1]||                dp[i][j-1]&&s1[j]==s3[i+j+1]

 

 1 public class Solution {

 2     public boolean isInterleave(String s1, String s2, String s3) {

 3         char c1[]=s1.toCharArray();

 4         

 5     char c2[]=s2.toCharArray();

 6     char c3[]=s3.toCharArray();

 7         int len1=s1.length();

 8         int len2=s2.length();

 9         int len3=s3.length();

10         if(len1+len2!=len3) return false;

11         if(len1==0) return s2.equals(s3);

12         if(len2==0) return s1.equals(s3);

13         

14         boolean dp[][]=new boolean[s1.length()+1][s2.length()+1];

15         dp[0][0]=true;

16         for(int i=1;i<=len1;i++)

17         {

18             

19             dp[i][0]=dp[i-1][0]&&c1[i-1]==c3[i-1];

20             

21         }

22         for(int j=1;j<=len2;j++)

23         {

24             

25             dp[0][j]=dp[0][j-1]&&c2[j-1]==c3[j-1];

26         }

27         

28         

29        

30         for(int i=1;i<=len1;i++)

31         {

32             

33             for(int j=1;j<=len2;j++)

34             {

35                 

36                 

37                 dp[i][j]=dp[i-1][j]&&(c1[i-1]==c3[i+j-1]);

38                 if(dp[i][j]) continue;

39                   dp[i][j]=dp[i][j-1]&&(c2[j-1]==c3[i+j-1]);

40             

41             }

42             

43             

44             

45         }

46          

47 

48          return dp[len1][len2];

49         

50         

51     }

52     

53 }

View Code