leetcode 日经贴，Cpp code -Scramble String-优快云博客

本文链接：https://blog.youkuaiyun.com/bachelorchen/article/details/44338883

本文介绍了一种判断两个字符串是否可以通过乱序操作相互转换的算法。该算法通过递归检查子串并比较字符频率来确定两个字符串是否为乱序字符串。文章详细展示了如何通过从左到右和从右到左两种方式检查字符串匹配情况。

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class Solution {
public:
    bool isScramble2(string s1, string s2) {
        int n = s1.length();
        if (n <= 3) return true;
        //from left
        int match = 0;
        for (int i = 0; i < n; ++i) {
            ++count[s1[i]];
            if(count[s1[i]] == 0) --match;
            else if(count[s1[i]] == 1) ++match;
            --count[s2[i]];
            if(count[s2[i]] == 0) --match;
            else if(count[s2[i]] == -1) ++match;
            if (match == 0 && i < n - 1) {
                if (isScramble2(s1.substr(0, i + 1), s2.substr(0, i + 1)) &&
                    isScramble2(s1.substr(i + 1, n - i - 1), s2.substr(i + 1, n - i - 1)))
                    return true;
                break;
            }
        }
        //from right
        for (int i = 0; i < n; ++i) {
            ++count[s1[i]];
            if(count[s1[i]] == 0) --match;
            else if(count[s1[i]] == 1) ++match;
            --count[s2[n - 1 - i]];
            if(count[s2[n - 1 - i]] == 0) --match;
            else if(count[s2[n - 1 - i]] == -1) ++match;
            if (match == 0 && i < n - 1) {
                if (isScramble2(s1.substr(0, i + 1), s2.substr(n - 1 - i, i + 1)) &&
                    isScramble2(s1.substr(i + 1, n - i - 1), s2.substr(0, n - i - 1)))
                    return true;
                break;
            }
        }
        return false;
    }
    bool isScramble(string s1, string s2) {
        if (s1.length() != s2.length())
            return false;
        memset(count, 0, sizeof(count));
        string ss1 = s1, ss2 = s2;
        sort(ss1.begin(), ss1.end());
        sort(ss2.begin(), ss2.end());
        for (int i = 0; i < s1.length(); ++i) {
            if (ss1[i] != ss2[i]) return false;
        }
        return isScramble2(s1, s2);
    }
private:
    int count[256];
};