力扣算法题—097交错字符串

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

//使用动态规划
//  Ø d b b c a
//Ø T F F F F F
//a T F F F F F
//a T T T T T F
//b F T T F T F
//c F F T T T T
//c F F F T F T

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if (s1.size() + s2.size() != s3.size()) return false;
        int n1 = s1.size(), n2 = s2.size();
        vector<vector<bool>> dp(n1 + 1, vector<bool>(n2 + 1, false));
        for (int i = 0; i <= n1; ++i) {
            for (int j = 0; j <= n2; ++j) {
                if (i == 0 && j == 0) {
                    dp[i][j] = true;
                }
                else if (i == 0) {
                    dp[i][j] = dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];
                }
                else if (j == 0) {
                    dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1];
                }
                else {
                    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]);
                }
            }
        }
        return dp[n1][n2];
    }
};

//使用带优化的DFS来做
class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if (s1.size() + s2.size() != s3.size()) return false;
        unordered_set<int> s;
        return helper(s1, 0, s2, 0, s3, 0, s);
    }
    bool helper(string& s1, int i, string& s2, int j, string& s3, int k, unordered_set<int>& s) {
        int key = i * s3.size() + j;
        if (s.count(key)) return false;
        if (i == s1.size()) return s2.substr(j) == s3.substr(k);
        if (j == s2.size()) return s1.substr(i) == s3.substr(k);
        if ((s1[i] == s3[k] && helper(s1, i + 1, s2, j, s3, k + 1, s)) ||
            (s2[j] == s3[k] && helper(s1, i, s2, j + 1, s3, k + 1, s))) return true;
        s.insert(key);
        return false;
    }
};

//使用BFS
class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if (s1.size() + s2.size() != s3.size()) return false;
        int n1 = s1.size(), n2 = s2.size(), n3 = s3.size(), k = 0;
        unordered_set<int> s;
        queue<int> q{ {0} };
        while (!q.empty() && k < n3) {
            int len = q.size();
            for (int t = 0; t < len; ++t) {
                int i = q.front() / n3, j = q.front() % n3; q.pop();
                if (i < n1 && s1[i] == s3[k]) {
                    int key = (i + 1) * n3 + j;
                    if (!s.count(key)) {
                        s.insert(key);
                        q.push(key);
                    }
                }
                if (j < n2 && s2[j] == s3[k]) {
                    int key = i * n3 + j + 1;
                    if (!s.count(key)) {
                        s.insert(key);
                        q.push(key);
                    }
                }
            }
            ++k;
        }
        return !q.empty() && k == n3;
    }
};