[hard]97. Interleaving String

交错字符串判断

原创于 2016-11-08 14:23:32 发布 · 234 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #DP

本文介绍了一个简单的动态规划解决方案来判断字符串s3是否可以通过交错字符串s1和s2形成。通过三维布尔数组跟踪所有可能的组合状态，实现了从初始状态逐步推导出最终结果的过程。

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.

solution:

Not thing to say. A very very simple and naive raw Dynamic Programming. Why it is in hard section?

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        bool f[s1.size() + 1][s2.size() + 1][s3.size() + 1];
        memset(f, 0, sizeof(f));
        f[0][0][0] = true;
        for (int i = 0; i <= s1.size(); ++ i)
            for (int j = 0; j <= s2.size(); ++ j)
                for (int k = 0; k < s3.size(); ++ k) if (f[i][j][k]) {
                    if (i < s1.size() && s1[i] == s3[k])
                        f[i + 1][j][k + 1] = true;
                    if (j < s2.size() && s2[j] == s3[k])
                        f[i][j + 1][k + 1] = true;
                }
        return f[s1.size()][s2.size()][s3.size()];
    }
};