Interleaving String

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.

算法：使用动态规划，f[i][j]表示s1的前i个字符和s2的前j个字符是否能组成s3的前i+j个字符，f[i][i]=true，条件：f[i-1][j]=true&&s1[i-1]==s3[i+j-1]或者f[i][j-1]=true&&s2[j-1]=s3[i+j-1] 不满足条件则：f[i][j]=false;最后f[s1.size()][s2.size()]即为所求，代码如下：（一开始被各种下标绕晕了，想了半天终于想明白了）

 1 class Solution {
 2 public:
 3     bool isInterleave(string s1, string s2, string s3) {
 4       int len1=s1.size();
 5       int len2=s2.size();
 6       int len3=s3.size();
 7       if(len1+len2!=len3)  return false;
 8       bool f[len1+1][len2+1];
 9       f[0][0]=true;
10       for(int i=1;i<=len1;i++)
11       {
12           if(f[i-1][0]&&s1[i-1]==s3[i-1])  f[i][0]=true;
13           else f[i][0]=false;
14       }
15       for(int i=1;i<len2+1;i++)
16       {
17           if(f[0][i-1]&&s2[i-1]==s3[i-1])  f[0][i]=true;
18           else f[0][i]=false;
19       }
20
21       for(int i=1;i<=len1;i++)
22       {
23           for(int j=1;j<=len2;j++)
24           {
25               if(s1[i-1]==s3[i+j-1]&&f[i-1][j]||s2[j-1]==s3[i+j-1]&&f[i][j-1])  f[i][j]=true;
26               else f[i][j]=false;
27           }
28       }
29       return f[len1][len2];
30     }
31 };

转载于:https://www.cnblogs.com/sqxw/p/4007977.html

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