https://leetcode.com/problems/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.
这题有点类似于之前的求编辑距离的题：http://www.cnblogs.com/biyeymyhjob/archive/2012/09/28/2707343.html
定义f[i][j]为 以s1[0，i]s2[0，j]匹配s3[i][j]
当j==0时，f[i][0]= f[i-1][0] && s1[i-1]==s3[i-1]
当i==0时，f[0][j]= f[0][j-1] && s2[j-1]==s3[j-1]
当i>=1&&j>=1时，f[i][j]= (f[i-1][j] && s1[i-1]==s3[i-1][j]) || (f[i][j-1] && s2[j-1]== s3[i][j-1])
bool isInterleave(string s1, string s2, string s3) {
if(s3.length()!=s1.length()+s2.length())
return false;
vector<vector<bool>> f(s1.length()+1,vector<bool>(s2.length()+1,true));
for(int i=1;i<=s1.length();i++)
f[i][0]= f[i-1][0] && s1[i-1]==s3[i-1];
for(int i=1;i<=s2.length();i++)
f[0][i]=f[0][i-1] && s2[i-1]==s3[i-1];
for(int i=1;i<=s1.length();i++)
{
for(int j=1;j<=s2.length();j++)
{
f[i][j]=(f[i-1][j] && s1[i-1]==s3[i+j-1]) || (f[i][j-1] && s2[j-1]==s3[i+j-1]);
}
}
return f[s1.length()][s2.length()];
}
原文：http://www.cnblogs.com/573177885qq/p/6236113.html