97. Interleaving String（dp）

题目：

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.

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深搜：

超时

public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if(s1.length()+s2.length()!=s3.length())return false;
return dfs(s1,s2,s3,0,0,0);
}

private boolean dfs(String s1,String s2,String s3, int l1,int l2,int l3)
{
if(l3 == s3.length())return true;
if(s1.length()>l1 && s1.charAt(l1)==s3.charAt(l3))
{
if(dfs(s1,s2,s3,l1+1,l2,l3+1))
{
return true;
}
}
if(s2.length()>l2 && s2.charAt(l2)==s3.charAt(l3))
{
if(dfs(s1,s2,s3,l1,l2+1,l3+1))
{
return true;
}
}
return false;

}
}

动态规划：
dp[i][j]表示s1的前i个和s2的前j个能构成s3的前i+j个

动态规划方程：dp[i][j] = dp[i - 1][j] && s2[i - 1] == s3[i + j - 1]  ||  dp[i][j - 1] && s1[i][j - 1] == s3[i + j - 1] 

public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if(s1.length()+s2.length()!=s3.length())return false;
boolean dp[][] = new boolean[s1.length()+1][s2.length()+1];
dp[0][0]=true;
for(int i=1;i<s1.length()+1;i++)
{
if(s1.charAt(i-1)==s3.charAt(i-1) && dp[i-1][0])
{
dp[i][0]=true;
}
}
for(int i=1;i<s2.length()+1;i++)
{
if(s2.charAt(i-1) == s3.charAt(i-1)&& dp[0][i-1])
{
dp[0][i]=true;
}
}

for(int i=1;i<s1.length()+1;i++)
{
for(int j=1;j<s2.length()+1;j++)
{
dp[i][j] = (s1.charAt(i-1)==s3.charAt(i+j-1) && dp[i-1][j])||(s2.charAt(j-1)==s3.charAt(i+j-1) && dp[i][j-1]);
}
}
return dp[s1.length()][s2.length()];
}
}