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 public class Solution { 2 public static void main(String[] args) 3 { 4 Solution s=new Solution(); 5 String s1="ab"; 6 String s2=""; 7 String s3="ba"; 8 System.out.println(s.isInterleave(s1, s2, s3)); 9 } 10 11 public boolean isInterleave(String s1, String s2, String s3) { 12 int len1=s1.length(); 13 int len2=s2.length(); 14 int len3=s3.length(); 15 16 if (s3==null||s3.length()==0) 17 return false; 18 19 if (len3!=len1+len2) 20 return false; 21 22 char[] ch1=s1.toCharArray(); 23 char[] ch2=s2.toCharArray(); 24 char[] ch3=s3.toCharArray(); 25 26 Boolean flag=isInterleave(ch1,len1-1,ch2,len2-1,ch3,len3-1); 27 return flag; 28 } 29 30 public boolean isInterleave(char[] ch1,int len1,char[] ch2,int len2, char[] ch3,int len3) 31 { 32 Boolean result=false; 33 char c1=' ',c2=' ',c3=' '; 34 if(len1>=0) 35 c1=ch1[len1]; 36 37 if(len2>=0) 38 c2=ch2[len2]; 39 40 if(len3>=0) 41 c3=ch3[len3]; 42 43 if (c3!=c1&&c3!=c2) 44 return false; 45 46 if(len3==1&&len1==1&&c3==c1) 47 { 48 return true; 49 } 50 51 if(len3==1&&len2==1&&c3==c2) 52 { 53 return true; 54 } 55 56 if (c3==c1&&c3!=c2) 57 { 58 result=isInterleave(ch1,len1-1,ch2,len2,ch3,len3-1); 59 } 60 61 if (c3==c2&&c3!=c1) 62 { 63 result=isInterleave(ch1,len1,ch2,len2-1,ch3,len3-1); 64 } 65 66 if(c3==c2&&c3==c1) 67 { 68 result=isInterleave(ch1,len1-1,ch2,len2,ch3,len3-1); 69 if(result==false) 70 result=isInterleave(ch1,len1,ch2,len2-1,ch3,len3-1); 71 } 72 73 return result; 74 75 76 } 77 }
自己写的代码，在本地调没问题，但是提交到牛客网中显示运行错误，不知道怎么回事。（在牛客网提交时注意将main函数注释掉）