[算法作业-动态规划][LeetCode] 97. Interleaving String

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最新推荐文章于 2023-11-23 18:09:02 发布

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分类专栏： LeetCode解题文章标签： leetcode 算法动态规划

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本文介绍了一种使用动态规划解决交错字符串问题的方法。通过构建二维动态规划表dp，判断字符串s3是否能由s1和s2交错组成。文章详细解释了状态转移方程，并给出完整的C++代码实现。

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[算法作业-动态规划][LeetCode] 97. Interleaving String

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

For example,
Given:
s1 = “aabcc”,
s2 = “dbbca”,

When s3 = “aadbbcbcac”, return true.
When s3 = “aadbbbaccc”, return false.

dp[i][j] 表示s3[0:i+j] 是否可由 s1[0:i] 和 s2[0:j]构成。

dp[i][j] = (
dp[i][j]
|| (dp[i-1][j] && (s1[i-1] == s3[i+j-1]))
|| (dp[i][j-1] && (s2[j-1] == s3[i+j-1]))
)

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int len1 = s1.length();
        int len2 = s2.length();
        int len3 = s3.length();
        if (len1 + len2 != len3) return false;

        vector<vector<bool>> dp(len1+1, vector<bool>(len2+1, false));
        dp[0][0] = true;
        for (int i=1; i<=len1; ++i) {
            dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1]);
        }
        for (int j=1; j<=len2; ++j) {
            dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1]);
        }

        for (int i=1; i<=len1; ++i) {
            for (int j=1; j<=len2; ++j) {
                dp[i][j] = dp[i][j] || (dp[i-1][j] && (s1[i-1] == s3[i+j-1]));
                dp[i][j] = dp[i][j] || (dp[i][j-1] && (s2[j-1] == s3[i+j-1]));
            }
        }
        return dp[len1][len2];
    }
};