# 交错字符串

<p>给定三个字符串 <code>s1</code>、<code>s2</code>、<code>s3</code>，请你帮忙验证 <code>s3</code> 是否是由 <code>s1</code> 和 <code>s2</code><em> </em><strong>交错 </strong>组成的。</p><p>两个字符串 <code>s</code> 和 <code>t</code> <strong>交错</strong> 的定义与过程如下，其中每个字符串都会被分割成若干 <strong>非空</strong> 子字符串：</p><ul>	<li><code>s = s<sub>1</sub> + s<sub>2</sub> + ... + s<sub>n</sub></code></li>	<li><code>t = t<sub>1</sub> + t<sub>2</sub> + ... + t<sub>m</sub></code></li>	<li><code>|n - m| <= 1</code></li>	<li><strong>交错</strong> 是 <code>s<sub>1</sub> + t<sub>1</sub> + s<sub>2</sub> + t<sub>2</sub> + s<sub>3</sub> + t<sub>3</sub> + ...</code> 或者 <code>t<sub>1</sub> + s<sub>1</sub> + t<sub>2</sub> + s<sub>2</sub> + t<sub>3</sub> + s<sub>3</sub> + ...</code></li></ul><p><strong>提示：</strong><code>a + b</code> 意味着字符串 <code>a</code> 和 <code>b</code> 连接。</p><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0097.Interleaving%20String/images/interleave.jpg" style="width: 561px; height: 203px;" /><pre><strong>输入：</strong>s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"<strong><br />输出：</strong>true</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"<strong><br />输出：</strong>false</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>s1 = "", s2 = "", s3 = ""<strong><br />输出：</strong>true</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>0 <= s1.length, s2.length <= 100</code></li>	<li><code>0 <= s3.length <= 200</code></li>	<li><code>s1</code>、<code>s2</code>、和 <code>s3</code> 都由小写英文字母组成</li></ul>

## template

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

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```