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

func isInterleave(s1 string, s2 string, s3 string) bool {
    m, n := len(s1), len(s2)
    if m + n != len(s3) { //長(zhǎng)度不等肯定不能組成
        return false
    }
    dp := make([][]string, m+1) //dp[i][j]含義為s1的前i個(gè)字符，s2前j個(gè)字符，能否組成s3前i+j個(gè)字符，能的話填寫字符串
    for i:=0; i<=m; i++ {
        dp[i] = make([]string, n+1)
    }
    for i:=1; i<=m; i++ { //s2字符串為空的情況
        if s1[:i] == s3[:i] {
            dp[i][0] = s1[:i]
        }
    }
    for i:=1; i<=n; i++ { //s1字符串為空的情況
        if s2[:i] == s3[:i] {
            dp[0][i] = s2[:i]
        }
    }
    for i:=1; i<=m; i++ {
        for j:=1; j<=n; j++ {
            if dp[i-1][j] != "" && s1[i-1] == s3[i+j-1] { //上方字符串符合并且s1當(dāng)前字符和s3字符相等，
                dp[i][j] = dp[i-1][j] + string(s1[i-1])   //則當(dāng)前狀態(tài)等于上方字符串+s1當(dāng)前字符
            }
            if dp[i][j-1] != "" && s2[j-1] == s3[i+j-1] { //左側(cè)字符串符合并且s2當(dāng)前字符和s3字符相等
                dp[i][j] = dp[i][j-1] + string(s2[j-1])   //則當(dāng)前狀態(tài)等于左側(cè)字符串+s2當(dāng)前字符
            }
        }
    }
    if dp[m][n] == s3 { //右下角字符串是否等于s3，等于則能合并出，否則不能
        return true
    }
    return false
}