LeetCode——Interleaving String

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本文介绍了一种用于判断字符串s3是否能由字符串s1和s2交错组成的有效算法。通过三维动态规划方法，该算法能够高效地解决此问题，并提供了一个C++实现示例。

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.

» Solve this problem

设f[k][i][j]表示s1[0..i-1]加s2[0..j-1]能否拼成s3[0..k-1]，满足：i + j = k。

当f[k - 1][i][j - 1] == true && s2[j - 1] == s3[k - 1]时，f[k][i][j] = true；

或者

当f[k - 1][i - 1][j] == true && s1[i - 1] == s3[k - 1]时，f[k][i][j] = true。

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if (l1 + l2 != l3) {
            return false;
        }
        bool f[l3 + 1][l1 + 1][l2 + 1];
        memset(f, false, sizeof(bool) * (l3 + 1) * (l1 + 1) * (l2 + 1));
        f[0][0][0] = true;
        for (int k = 1; k <= l3; k++) {
            for (int i = 0; i <= k && i <= l1; i++) {
                int j = k - i;
                if (j > l2) {
                    continue;
                }
                if (i != 0 && f[k - 1][i - 1][j] && s1[i - 1] == s3[k - 1]) {
                    f[k][i][j] = true;
                }
                if (j != 0 && f[k - 1][i][j - 1] && s2[j - 1] == s3[k - 1]) {
                    f[k][i][j] = true;
                }
            }
        }
        return f[l3][l1][l2];
    }
};