0097. Interleaving String (M)

Interleaving String (M)

题目

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where they are divided into non-empty substrings such that:

• s = s1 + s2 + ... + sn
• t = t1 + t2 + ... + tm
• |n - m| <= 1
• The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

Constraints:

• 0 <= s1.length, s2.length <= 100
• 0 <= s3.length <= 200
• s1, s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

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题意

判断字符串s1和s2能否通过间隔插入的方法得到s3。

思路

动态规划。dp[i][j]为布尔值，表示s1.substring(0,i)和s2.substring(0,j)能否通过间隔插入组成s3.substring(0, i + j)。

Follow up可以用滚动数组优化到O(s2.length)。

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代码实现

Java

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s3.length() != s1.length() + s2.length()) return false;      
       
        boolean[][] dp = new boolean[s1.length() + 1][s2.length() + 1];

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

        return dp[s1.length()][s2.length()];
    }
}