Interleaving String leetcode

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.

定义boolean 数组result[][]表示s1的前i个字符和s2的前j个字符是否能交替组成s3的前i＋j个字符。

function：result[i][j] =   result[i-1][j] &&(s1[i-1] == s3[i+j-1]) ||  result[i][j -1] &&(s2[j-1] == s3[i+j-1])

initialize:

result[i][0] = (s1[0...i-1] == s3[0...i-1])

result[0][j] = (s2[0...i-1] == s3[0...i-1])

返回值： result[s1.length()][s2.length()]

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1.length() + s2.length() != s3.length()) {
            return false;
        }
        int m = s1.length();
        int n = s2.length();
        boolean[][] result = new boolean[m + 1][n + 1];
        result[0][0] = true;

        for (int i = 1; i < m + 1; i++) {
            if (result[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1)){
                result[i][0] = true;
            }
        }
        for (int j = 1; j < n + 1; j++) {
            if (result[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j -1)){
                result[0][j] = true;
            }
        }
        for (int i = 1; i < m + 1; i++) {
            for (int j = 1; j < n + 1; j++) {
                if (result[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) || result[i][j - 1] && s2.charAt(j -1 ) == s3.charAt(i + j - 1)){
                    result[i][j] = true;
                }
            }
        }
        return result[m][n];
    }
}