LeetCode 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.

 

1.  Naive Solution 

　　　　直接递归实现。

　　　　

bool isInterleave(string s1, string s2, string s3) {
    int n1 = (int)s1.length();
    int n2 = (int)s2.length();
    int n3 = (int)s3.length();
    
    if (n1 + n2 != n3) {
        return false;
    }
    if (n1 != 0 || n1 != 0) {
        if (n1 && n2) {
            if (s1[0] == s2[0] && s1[0] == s3[0]) {
                return isInterleave(s1.substr(1), s2, s3.substr(1)) || isInterleave(s1, s2.substr(1), s3.substr(1));
            }
            else if (s1[0] == s3[0])
                return isInterleave(s1.substr(1), s2, s3.substr(1));
            else
                return isInterleave(s1, s2.substr(1), s3.substr(1));
        }
        else if (n1) {
            return s1 == s3;
        }
        else {
            return s2 == s3;
        }
    }
    
    return false;
}

结果可想而知TLE，无法AC。考虑采用迭代重写

bool isInterleave(string s1, string s2, string s3) {
    int n1 = (int)s1.length();
    int n2 = (int)s2.length();
    int n3 = (int)s3.length();
    if (n1 + n2 != n3) return false;
    int i = 0;
    int j = 0;
    
    bool flag = false;
    stack<int> __S1;
    stack<int> __S2;
    stack<int> __S3;
    
    for (int k = 0; k < n3; ++k) {
        if ((i < n1 && s3[k] == s1[i]) && (j < n2 && s3[k] == s2[j])) {
            if (!flag) {
                __S1.push(i);
                __S2.push(j);
                __S3.push(k);

                ++i;
            }
            else {
                flag = false;
                ++j;
            }
        }
        else if (i < n1 && s3[k] == s1[i]) {
            ++i;
        }
        else if (j < n2 && s3[k] == s2[j]){
            ++j;
        }
        else {
            if (!__S3.empty()) {
                k = __S3.top();
                --k;
                i = __S1.top();
                j = __S2.top();
                __S3.pop();
                __S2.pop();
                __S1.pop();
                flag  = true;
            }
            else return false;
           
        }
    }
    
    return i == n1 && j == n2;
}

结果对于大集合时间缩短明显，但算法复杂度太高O(2^(N+M)), N为s1的长度，M为s2的长度。考虑Dynamic Programming以减少冗余的计算。

　　　　2. Advanced Solution

bool isInterleave(string s1, string s2, string s3) {
    int n1 = (int)s1.length();
    int n2 = (int)s2.length();
    int n3 = (int)s3.length();
    if (n1 + n2 != n3) return false;
    
    if (n1 == 0 || n2 == 0) {
        if (n1 == 0) {
            return s2 == s3;
        }
        else return s1 == s3;
    }
    
    vector<vector<int> > M(n1 + 1, vector<int>(n2 + 1, 0));
    M[0][0] = 1;
    
    for (int i = 1; i <= n1; ++i) {
        if (s1[i-1] == s3[i-1]) {
            M[i][0] = 1;
        }
        else break;
    }
    
    for (int j = 1; j <= n2; ++j) {
        if (s2[j-1] == s3[j-1]) {
            M[0][j] = 1;
        }
        else break;
    }

    for (int i = 1; i <= n1; ++i) {
        for (int j = 1; j <= n2; ++j) {
            if (M[i-1][j] && s1[i-1] == s3[i+j-1]) {
                M[i][j] = 1;
            }
            else if (M[i][j-1] && s2[j-1] == s3[i+j-1]) {
                M[i][j] = 1;
            }
            else M[i][j] = 0;
        }
        
    }
    
    return M[n1][n2];
}

该算法基于以下考虑，若s3.substr(0, i+j)是s1.substr(0, i) 与s2.substr(0, j)的 interleaving string 那么只有两种可能

1. s1[i] == s3[i+j-1] 并且 s3.substr(0, i+j-1)是s1.substr(0, i-1) 与 s2.substr(0, j)的 interleaving string
2. s2[j] == s3[i+j-1] 并且 s3.substr(0, i+j-1)是s1.substr(0, i)与s2.substr(0, j-1)的 interleaving string