Q97 交错字符串

最新推荐文章于 2025-06-13 20:08:32 发布

转载最新推荐文章于 2025-06-13 20:08:32 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/WeichengDDD/p/10801861.html

博客主要围绕验证字符串展开，给定三个字符串 s1、s2、s3，需验证 s3 是否由 s1 和 s2 交错组成，还给出了示例，并标注了转载来源。

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给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。

示例 1:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true

示例 2:

输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1 == null || s2 == null || s3 == null || s3.length() != (s1.length() + s2.length()))
            return false;
        else if (s1.equals("") && s2.equals(s3) || s2.equals("") && s1.equals(s3))
            return true;

        int[][] dp = new int[s1.length() + 1][s2.length() + 1];
        char[] chs1 = s1.toCharArray();
        char[] chs2 = s2.toCharArray();
        char[] chs3 = s3.toCharArray();

        for (int i = 1; i <= chs1.length; i++) {
            if (chs1[i - 1] == chs3[i - 1])
                dp[i][0] = dp[i - 1][0] + 1;
            else {
                int t = dp[i - 1][0];
                while(i < chs1.length) {
                    dp[i - 1][0] = t;
                    i++;
                }
            }
        }

        for (int i = 1; i <= chs2.length; i++) {
            if (chs2[i - 1] == chs3[i - 1])
                dp[0][i] = dp[0][i - 1] + 1;
            else {
                int t = dp[0][i - 1];
                while(i < chs2.length) {
                    dp[0][i - 1] = t;
                    i++;
                }
            }
        }

        for (int i = 1; i <= chs1.length; i++) {
            for (int j = 1; j <= chs2.length; j++) {
                int ti = dp[i - 1][j];
                int tj = dp[i][j - 1];

                int t1 = ti + (chs1[i - 1] == chs3[ti] ? 1 : 0);
                int t2 = tj + (chs2[j - 1] == chs3[tj] ? 1 : 0);

                dp[i][j] = t1 > t2 ? t1 : t2;
            }
        }

        return dp[chs1.length][chs2.length] == chs3.length;
    }
}

转载于:https://www.cnblogs.com/WeichengDDD/p/10801861.html