97. 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.

s1 s2 内部字母相对顺序不变是否可以混合成s3

二维动态规划，s3中的第i+j个字符一定是由s1中的前i个字符和s2中的前j个字符组合而成第i+j个字符可能是s1中的第i个也可能是s2中的第j个

也可以从后往前分析，s3中的最后一个字符一定是s1或者s2的最后一个字符，假如是s2 那么对应位置为true，s2的索引往前移动一位，s3的索引向前移动一位继续遍历

hollyshit leetcode检查速度太慢一直submit failed，先把代码放这思路大致是这样的

可注意if的条件要有 dp[i][j-1] dp[i-1][j]==true 以及对长度这个基本条件的判断

public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if((s1==null&&(s2==s3))||(s2==null&&(s1==s3))) return true;
int len1=s1.length();
int len2=s2.length();
if(len1+len2!=s3.length()) return false;
boolean dp[][]=new boolean[len1+1][len2+1];
dp[0][0]=true;
for(int i=0;i<=len1;i++){
for(int j=0;j<=len2;j++){
if(dp[i][j]==true
||(i>=1&&dp[i-1][j]==true&&(s1.charAt(i-1)==s3.charAt(i+j-1)))
||(j>=1&&dp[i][j-1]==true&&(s2.charAt(j-1)==s3.charAt(i+j-1))))
dp[i][j]=true;
else dp[i][j]=false;
}
}
return dp[len1][len2];
}
}

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