题目连接:Leetcode 087 Scramble String
解题思路:DFS,每次枚举分割点,判断左右子串是否是Scramble String。不加剪枝的话是会超时,在每次递归判断前,判断当前两个字符串字符个数是否一致,不一致的话直接返回false。
class Solution { public: bool isScramble(string s1, string s2) { if (s1.size() != s2.size()) return false; if (s1 == s2) return true; int n = s1.size(); int s1Map[128] = {0}, s2Map[128] = {0}; for (int i = 0; i < n; i++) { s1Map[s1[i]]++; s2Map[s2[i]]++; } for (int i = 0; i < 128; i++) if(s1Map[i] != s2Map[i]) return false; for (int i = 1; i < n; i++) { if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)) ) return true; if (isScramble(s1.substr(0, i), s2.substr(n-i)) && isScramble(s1.substr(i), s2.substr(0, n-i)) ) return true; } return false; } };