leetCode练习（97）

题目：Interleaving String          

难度：hard

问题描述：

Given s1, s2, s3, find whether s3 is formed by the interleaving ofs1 and s2.

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

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

解题思路：判断s3是否由s1、s2顺序插入形成。

一开始使用了DFS回溯法，很简单，结果超时。这里放上代码：

public static boolean isInterleave(String s1, String s2, String s3) {if(s3.equals("")){if(s1.equals("")&&s2.equals("")){return true;}else{return false;}}if(s1.equals("")){return s2.equals(s3);}if(s2.equals("")){return s1.equals(s3);}char c1=s1.charAt(0);char c2=s2.charAt(0);char c3=s3.charAt(0);if(c3==c1){boolean temp=isInterleave(s1.substring(1),s2,s3.substring(1));if(c3==c2){return temp||isInterleave(s1,s2.substring(1),s3.substring(1));}else{return temp;}}else{if(c3==c2){return isInterleave(s1,s2.substring(1),s3.substring(1));}else{return false;}}    }

想想也是，字符串长度太大，DFS指数增长太可怕。这时看到提示使用动态规划。想到了可以使用二维动态规划。

s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbcbcac"

对于Table[I][j],若table[I-1][j]=1且s2[j]=s3[I+1-1];或者table[I][j-1]=1且s1[I]=s3[I+j-1] ,那么Table[I][j]=1，否则=0；

具体代码如下：

public static boolean isInterleave2(String s1, String s2, String s3) {if(!(s1.length()+s2.length()==s3.length())){return false;}if(s1.equals("")||s2.equals("")){return s3.equals(s1+s2);}if(s3.equals("")){return false;}boolean[] row=new boolean[s1.length()];boolean[] column=new boolean[s2.length()];for(int i=0;i<row.length;i++){if(s1.charAt(i)==s3.charAt(i))row[i]=true;else break;}for(int i=0;i<column.length;i++){if(s2.charAt(i)==s3.charAt(i))column[i]=true;else break;}for(int i=0;i<row.length;i++){for(int j=0;j<column.length;j++){if((row[i]&&s2.charAt(j)==s3.charAt(i+j+1))||(column[j]&&s1.charAt(i)==s3.charAt(i+j+1))){column[j]=true;row[i]=true;}else{column[j]=false;row[i]=false;}}}return column[s2.length()-1];}