C++扰乱字符串

如题：我们可以采用动态规划解决问题：


上面定义中k对应代码中的len，w对应了代码中的k，因为代码比较长，换了变量定义方便我们看懂

class Solution {
public:bool isScramble(string s1, string s2) {int n = s1.size();int m = s2.size();// 如果长度不相等，则必然不相等if (n != m){return false;}// 如果字符不相同，那么必然也不可能相等int cnt1[26];int cnt2[26];memset(cnt1, 0, sizeof(int)*26);memset(cnt2, 0, sizeof(int)*26);for (int i = 0; i < n; ++i){++cnt1[s1[i]-'a'];++cnt2[s2[i]-'a'];}for (int i = 0; i < 26; ++i){if (cnt1[i] != cnt2[i]){return false;}}// 这里额外多取一个是想覆盖len=n的情况，而无需额外去减1vector<vector<vector<int>>> d(n, vector<vector<int>>(n, vector<int>(n+1, 0)));for (int i = 0; i < n; ++i){for (int j = 0; j < n; ++j){d[i][j][1] = s1[i] == s2[j];}}// 这里长度覆盖到n位置for (int len = 2; len <= n; ++len){for (int i = 0; i <= n - len; ++i){for (int j = 0; j <= n - len; ++j){// 内部子长度去尝试是否满足条件, 这里至少要保证分段大于0，所以范围是1~len-1for (int k = 1; k <= len-1; ++k){if (d[i][j][k] && d[i+k][j+k][len-k]){d[i][j][len] = true;// cout << i << "," << j << "," << len << " true" << endl;// 退出子长度k的循环break;}if (d[i][j+len-k][k] && d[i+k][j][len-k]){d[i][j][len] = true;// cout << i << "," << j << "," << len << " true" << endl;// 退出子长度k的循环break;}}}}}return d[0][0][n];}
};