count-the-repetitions

最新推荐文章于 2020-11-20 23:17:57 发布

转载最新推荐文章于 2020-11-20 23:17:57 发布 · 143 阅读

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#数据结构与算法

本文探讨了LeetCode上一道关于字符串重复计数的问题，并提供了几种不同的解决方案。其中包括一种使用额外数组进行优化的方法，以及一个更为简洁的遍历比较方法。尽管进行了优化，但在某些情况下仍会出现超时问题。

https://leetcode.com/problems/count-the-repetitions/

下面是我的方法，结果对的，超时了。。。

package com.company;


class Solution {
    public int getMaxRepetitions(String s1, int n1, String s2, int n2) {
        int len1 = s1.length();
        int[][]stores = new int[26][len1];
        int[] pos = new int[26];
        int[] cur = new int[26];
        int index;

        for (int i=0; i<len1; i++) {
            index = s1.charAt(i)-'a';
            stores[index][pos[index]] = i;
            pos[index] = pos[index] + 1;
        }

        int curPos = 0;
        int ret = 0;
        int len2 = s2.length();
        while (true) {

            for (int i=0; i<n2; i++) {
                for (int j=0; j<len2; j++) {
                    index = s2.charAt(j) - 'a';

                    if (cur[index] >= pos[index] * n1) {

                        return ret;
                    }

                    int newPos = 0;
                    do {
                        newPos = cur[index] / pos[index] * len1 + stores[index][cur[index] % pos[index]];
                        cur[index] = cur[index] + 1;
                    } while (newPos < curPos && cur[index] < pos[index] * n1);

                    if (newPos < curPos) {
                        return ret;
                    }
                    curPos = newPos + 1;


                }
            }
            ret++;

        }

    }
}

public class Main {

    public static void main(String[] args) throws InterruptedException {

        String s1 = "niconiconi";
        int n1 = 99981;
        String s2 = "nico";
        int n2 = 81;

        Solution solution = new Solution();
        int ret  = solution.getMaxRepetitions(s1, n1, s2, n2);

        // Your Codec object will be instantiated and called as such:
        System.out.printf("ret:%d\n", ret);

        System.out.println();

    }

}

优化之后的结果，还是超时：

加了string到array的优化，另外每次循环之后坐个判断剪枝。

package com.company;


class Solution {
    public int getMaxRepetitions(String s1, int n1, String s2, int n2) {
        int len1 = s1.length();
        int[][]stores = new int[26][len1];
        int[] pos = new int[26];
        int[] cur = new int[26];
        int index;

        for (int i=0; i<len1; i++) {
            index = s1.charAt(i)-'a';
            stores[index][pos[index]] = i;
            pos[index] = pos[index] + 1;
        }

        int curPos = 0;
        int ret = 0;
        int len2 = s2.length();
        char[] array2 = s2.toCharArray();
        while (true) {

            for (int i=0; i<n2; i++) {
                for (int j=0; j<len2; j++) {
                    index = array2[j] - 'a';

                    int newPos = 0;
                    while (cur[index] < pos[index] * n1) {
                        newPos = cur[index] / pos[index] * len1 + stores[index][cur[index] % pos[index]];
                        cur[index] = cur[index] + 1;
                        if (newPos >= curPos) {
                            break;
                        }
                    }

                    if (newPos < curPos) {
                        /*System.out.printf("index %d cur[index] %d pos[index] %d cur/-pos %d, store %d\n",
                                index, cur[index], pos[index], cur[index] % pos[index], stores[index][cur[index] % pos[index]]);

                        System.out.printf("newPos %d curPos %d\n",
                                newPos, curPos);
                                */
                        return ret;
                    }
                    curPos = newPos + 1;


                }
            }
            ret++;
            for (int i=0; i<26; i++) {
                if (pos[i] > 0 && cur[i] >= pos[i] * n1) {
                    return ret;
                }
            }

        }

    }
}

public class Main {

    public static void main(String[] args) throws InterruptedException {

        String s1 = "acb";
        int n1 = 4;
        String s2 = "ab";
        int n2 = 2;

        Solution solution = new Solution();
        int ret  = solution.getMaxRepetitions(s1, n1, s2, n2);

        // Your Codec object will be instantiated and called as such:
        System.out.printf("ret:%d\n", ret);

        System.out.println();

    }

}

用了这种Brute Force的方法，居然比我的快。。。。。。

public class Solution {
    public int getMaxRepetitions(String s1, int n1, String s2, int n2) {
        char[] array1 = s1.toCharArray(), array2 = s2.toCharArray();
        int count1 = 0, count2 = 0, i = 0, j = 0;
        
        while (count1 < n1) {
            if (array1[i] == array2[j]) {
                j++;
                if (j == array2.length) {
                    j = 0;
                    count2++;
                }
            }
            i++;
            if (i == array1.length) {
                i = 0;
                count1++;
            }
        }
        
        return count2 / n2;
    }
}