97. 交错字符串

给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下，其中每个字符串都会被分割成若干 非空 子字符串：

s = s1 + s2 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...
提示：a + b 意味着字符串 a 和 b 连接。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/interleaving-string
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

动态规划

import java.util.Scanner;

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {

        if (s1.length() + s2.length() != s3.length()) {
            return false;
        }

        if (s1.length() == 0) {
            return s2.equals(s3);
        }

        if (s2.length() == 0) {
            return s1.equals(s3);
        }

        int n = s1.length(), m = s2.length();
        boolean[][] dp = new boolean[n + 1][m + 1];

        dp[0][0] = true;

        for (int i = 1; i <= n; ++i) {
            dp[i][0] = dp[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1);
        }

        for (int i = 1; i <= m; ++i) {
            dp[0][i] = dp[0][i - 1] && s2.charAt(i - 1) == s3.charAt(i - 1);
        }

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

        return dp[n][m];
    }
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().isInterleave(in.next(), in.next(), in.next()));
        }
    }
}

动态规划（空间压缩）

import java.util.Scanner;

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {

        if (s1.length() + s2.length() != s3.length()) {
            return false;
        }

        if (s1.length() == 0) {
            return s2.equals(s3);
        }

        if (s2.length() == 0) {
            return s1.equals(s3);
        }

        if (s1.length() < s2.length()) {
            String t = s1;
            s1 = s2;
            s2 = t;
        }

        int n = s1.length(), m = s2.length();
        boolean[] dp = new boolean[m + 1];

        dp[0] = true;

        for (int i = 1; i <= m; ++i) {
            dp[i] = dp[i - 1] && s2.charAt(i - 1) == s3.charAt(i - 1);
        }

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

        return dp[m];
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().isInterleave(in.next(), in.next(), in.next()));
        }
    }
}