Java C++题解leetcode字符串轮转KMP算法详解

目录

• 题目要求
• 思路一：双指针（模拟）
• Java
• C++
• 思路二：子串
• 手写KMP
• Java
• dp
• C++
• dp
• 调API
• Java
• C++
• 总结

题目要求

思路一：双指针（模拟）

Java

class Solution {
    public boolean isFlipedString(String s1, String s2) {
        if (s1.length() != s2.length())
            return false;
        int n = s1.length();
        if (n == 0)
            return true;
        for (int i = 0; i < n; i++) {
            boolean res = true;
            for (int j = 0; j < n; j++) {
                if (s1.charAt((i + j) % n) != s2.charAt(j)) {
                    res = false;
                    break;
                }
            }
            if (res)
                return true;
        }
        return false;
    }
}

• 时间复杂度：O(n^2)
• 空间复杂度：O(1)

C++

class Solution {
public:
    bool isFlipedString(string s1, string s2) {
        if (s1.size() != s2.size())
            return false;
        int n = s1.size();
        if (n == 0)
            return true;
        for (int i = 0; i < n; i++) {
            bool res = true;
            for (int j = 0; j < n; j++) {
                if (s1[(i + j) % n] != s2[j]) {
                    res = false;
                    break;
                }
            }
            if (res)
                return true;
        }
        return false;
    }
};

• 时间复杂度：O(n^2)
• 空间复杂度：O(1)

思路二：子串

手写KMP

KMP的思路可以参考KMP 算法详解和详解KMP算法

Java

get_next

class Solution {
    public boolean isFlipedString(String s1, String s2) {
        if (s1.length() != s2.length())
            return false;
        int n = s1.length();
        if (n == 0)
            return true;
        int[] nxt = new int[n];
        nxt[0] = -1;
        int j = 0; // pat指针
        int k = -1; // 跳转位置
        while (j < n - 1) {
            if (k == -1 || s2.charAt(j) == s2.charAt(k)) {
                if (s2.charAt(++j) == s2.charAt(++k))
                    nxt[j] = nxt[k]; // 连续相同
                else
                    nxt[j] = k;
            }
            else
                k = nxt[k];
        }
        String txt = s1 + s1;
        j = 0;
        for (int i = 0; i < 2 * n; i++) {
            if (j < n && txt.charAt(i) == s2.charAt(j))
                j++;
            else {
                j = nxt[j];
                if (j == -1)
                    j++;
            }
            if (j == n)
                return true;
        }
        return false;
    }
}

dp

class Solution {
    public boolean isFlipedString(String s1, String s2) {
        if (s1.length() != s2.length())
            return false;
        int n = s1.length();
        if (n == 0)
            return true;
        int[][] dp = new int[n][256]; // dp[state][char] = nxt state
        dp[0][s2.charAt(0)] = 1;
        int x = 0; // 影子状态
        for (int s = 1; s < n; s++) {
            for (int c = 0; c < 256; c++)
                dp[s][c] = dp[x][c];
            dp[s][s2.charAt(s)] = s + 1;
            x = dp[x][s2.charAt(s)];
        }
        String txt = s1 + s1;
        int state = 0;
        for (int i = 0; i < 2 * n; i++) {
            state = dp[state][txt.charAt(i)];
            if (state == n)
                return true;
        }
        return false;
    }
}

• 时间复杂度：O(n)
• 空间复杂度：O(n)

C++

get_next

class Solution {
public:
    bool isFlipedString(string s1, string s2) {
        if (s1.size() != s2.size())
            return false;
        int n = s1.size();
        if (n == 0)
            return true;
        int nxt[n];
        nxt[0] = -1;
        int j = 0; // pat指针
        int k = -1; // 跳转位置
        while (j < n - 1) {
            if (k == -1 || s2[j] == s2[k]) {
                if (s2[++j] == s2[++k])
                    nxt[j] = nxt[k]; // 连续相同
                else
                    nxt[j] = k;
            }
            else
                k = nxt[k];
        }
        string txt = s1 + s1;
        j = 0;
        for (int i = 0; i < 2 * n; i++) {
            if (j < n && txt[i] == s2[j])
                j++;
            else {
                j = nxt[j];
                if (j == -1)
                    j++;
            }
            if (j == n)
                return true;
        }
        return false;
    }
};

dp

class Solution {
public:
    bool isFlipedString(string s1, string s2) {
        if (s1.size() != s2.size())
            return false;
        int n = s1.size();
        if (n == 0)
            return true;
        int dp[n][256]; // dp[state][char] = nxt state
        memset(dp, 0, sizeof(dp));
        dp[0][s2[0]] = 1;
        int x = 0; // 影子状态
        for (int s = 1; s < n; s++) {
            for (int c = 0; c < 256; c++)
                dp[s][c] = dp[x][c];
            dp[s][s2[s]] = s + 1;
            x = dp[x][s2[s]];
        }
        string txt = s1 + s1;
        int state = 0;
        for (int i = 0; i < 2 * n; i++) {
            state = dp[state][txt[i]];
            if (state == n)
                return true;
        }
        return false;
    }
};

• 时间复杂度：O(n)
• 空间复杂度：O(n)

调API

Java

class Solution {
    public boolean isFlipedString(String s1, String s2) {
        return s1.length() == s2.length() && (s1 + s1).contains(s2);
    }
}

• 时间复杂度：O(n)，取决于语言匹配子字符串机制
• 空间复杂度：O(n)

C++

class Solution {
public:
    bool isFlipedString(string s1, string s2) {
        return s1.size() == s2.size() && (s1 + s1).find(s2) != string::npos;
    }
};

• 时间复杂度：O(n)，取决于语言匹配子字符串机制
• 空间复杂度：O(n)

impl Solution {
    pub fn is_fliped_string(s1: String, s2: String) -> bool {
        s1.len() == s2.len() && s2.repeat(2).contains(&s1)
    }
}

• 时间复杂度：O(n)，取决于语言匹配子字符串机制
• 空间复杂度：O(n)

总结

做过了轮转的题就能很快意识到拼接然后找子串，本来调个API结束结果发现了KMP算法，就浅学一下~

时间耗在算法了就掠过rust各种辣~

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