[LeetCode] Scramble String

难得一次AC

 

1. 明显的区间动态规划。f[i][j][p][q] 表示s1[i..j]和s2[p..q]是否是scramble String，最后求f[0][n-1][0][n-1]即可。

2. 实际上永远有两段区间相等，为了节约内存，增强代码可读性，并且不牺牲时间复杂度，将状态存入一个state类，并用unordered_map来保存。

 

class Solution {
public:
    struct state{
        int x1, y1, x2, y2;
        state(int x1, int y1,int x2,int y2):x1(x1),y1(y1),x2(x2),y2(y2){}
        state():x1(0),x2(0),y1(0),y2(0){}
        
        bool operator==(const state& rhs) const {
            return x1 == rhs.x1 &&
            x2 == rhs.x2 &&
            y1 == rhs.y1 &&
            y2 == rhs.y2;
        }
    };
    
    struct StateHash {
        size_t operator()(const state& s) const {
            return std::hash<int>()(s.x1 + s.x2 * s.x2 + s.y1 * s.y1 * s.y1 + s.y2 * s.y2 * s.y2 * s.y2);
        }
    };
    
    bool isScramble(string s1, string s2) {
        unordered_map<state, int, StateHash> states;
        int n = s1.size();
        if (s2.size() != n) return false;
        state final_state(0,n-1,0,n-1);
        return dfs(s1, s2, final_state, states);
    }
    
    bool dfs(const string& s1, const string& s2, state s, unordered_map<state, int, StateHash>& states) {
        auto it = states.find(s);
        if (it != states.end()) return it->second;
        bool ans = false;
        if (s.x1 == s.y1) {
            if (s1[s.x1] == s2[s.x2]) ans = true;
        } else {
            for (int i = s.x1; i < s.y1; i++) {
                int len = i - s.x1;
                if (dfs(s1, s2, state(s.x1, i, s.x2, s.x2 + len), states) && dfs(s1, s2, state(i+1, s.y1, s.x2+len+1, s.y2), states)) {
                    ans = true;
                    break;
                }
                
                if (dfs(s1, s2, state(s.x1, i, s.y2-len, s.y2), states) && dfs(s1, s2, state(i+1,s.y1, s.x2, s.y2-len-1), states)) {
                    ans = true;
                    break;
                }
            }
        }
        
        return states[s] = ans;
    }
};