Leetcode 97. Interleaving String? Staggered character discriminant problem solving report

1 ideas for solving problems

My problem? is to refer to this blog: Portal

Specifically, this question is to let us judge whether the string S3 by the characters in S1 and S2? That is, the S1 of each character is kept in a relatively orderly situation randomly inserted into the S2 whether can get S3 (described here, S1 and S2 can be interchangeable)

First of all, the traditional recursive way will certainly time out, certainly not.

The approach here is to expand the S1 and S2 into a matrix, assuming that the length is n,m, then construct a n*m DP matrix and use the DP method to judge.

Specifically (from the original text):

When the S1 arrives at the first element, the S2 arrives at the J element:

The right one step on the map is s2[j-1] matching s3[i+j-1].

One step down on the map is s1[i-1] matching s3[i+j-1].

Example: s1= "AA", s2= "AB", s3= "Aaba". The Mark 1 is feasible. Finally returns to the lower right corner.

0  a  b

0 1 1 0

A 1 1 1

A 1 0 1 2 original question

Given S1, S2, S3, find whether S3 are formed by the interleaving of S1 and S2.

For example,
Given:
S1 = "AABCC",
s2 = "DBBCA", while

s3 = "AADBBCBCAC", return true.
When s3 = ' AADBBBACCC ', return false.

3 AC Solution

public class Solution {//public boolean Isinterleave (string s1, String s2, string s3) {int n=s1==null?0
        : S1.length ();
        int M=s2==null?0:s2.length ();
        int K=s3==null?0:s3.length ();
        if (n+m!= k) return false;
        char[] S1 = S1.tochararray ();
        char[] S2 = S2.tochararray ();
        char[] S3 = S3.tochararray ();
        Boolean map[][]=new boolean[n+1][m+1]; According to the method described above, the state for (int i=0;i<=n;i++) {for (int j=0;j<=m;j++) {if (i==0 &&amp) is judged. ;
                j==0) {Map[i][j] = true;
                else if (i==0) {map[i][j] = Map[i][j-1] & s3[i+j-1]==s2[j-1];
                else if (j==0) {map[i][j] = Map[i-1][j] & s3[i+j-1]==s1[i-1]; else {Map[i][j] = (Map[i][j-1] & S3[i+j-1]==s2[j-1]) |
                (Map[i-1][j] & s3[i+j-1]==s1[i-1]);
}} return map[n][m];

    }
}