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;
}
}

(完)

count-the-repetitions的更多相关文章

  1. CH5702 Count The Repetitions

    题意 5702 Count The Repetitions 0x50「动态规划」例题 描述 定义 conn(s,n) 为 n 个字符串 s 首尾相接形成的字符串,例如: conn("abc& ...

  2. [Swift]LeetCode466. 统计重复个数 | Count The Repetitions

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc&qu ...

  3. 466. Count The Repetitions

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc&qu ...

  4. [LeetCode] Count The Repetitions 计数重复个数

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc&qu ...

  5. Leetcode: Count The Repetitions

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc&qu ...

  6. CH5702 Count The Repetitions[倍增dp]

    http://contest-hunter.org:83/contest/0x50%E3%80%8C%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%E3%80%8D%E4%B ...

  7. 【leetcode 字符串】466. Count The Repetitions

    https://leetcode.com/problems/count-the-repetitions/description/ 找循环节 https://www.cnblogs.com/grandy ...

  8. 第七周 Leetcode 466. Count The Repetitions 倍增DP (HARD)

    Leetcode 466 直接给出DP方程 dp[i][k]=dp[i][k-1]+dp[(i+dp[i][k-1])%len1][k-1]; dp[i][k]表示从字符串s1的第i位开始匹配2^k个 ...

  9. [LeetCode] 466. Count The Repetitions 计数重复个数

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc&qu ...

  10. LeetCode466. Count The Repetitions

    题目链接 传送门 题意 定义一个特殊的串, 现在给出串S1和S2的参数, 问: S2最多可以多少个连接起来扔是S1的子序列, 求出这个最大值 解题思路 注意s1与S1的区别, 可以去看题目描述, 预处 ...

随机推荐

  1. android19以上和以下uri转路径的方法

    android 19以上 /** * 专为Android4.4以上设计的从Uri获取文件路径 */ public static String getPath(final Context context ...

  2. c# 回调委托

    using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...

  3. easyui的页面等待提示层,即mask

    /* * 使用方法: * 开启:MaskUtil.mask(); * 关闭:MaskUtil.unmask(); * * MaskUtil.mask('其它提示文字...'); */ var Mask ...

  4. Hiking 分类: 比赛 HDU 函数 2015-08-09 21:24 3人阅读 评论(0) 收藏

    Hiking Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Subm ...

  5. laravel 重写以及500错误

    http://www.golaravel.com/laravel/docs/5.1/ sudo chmod 755 -R laravel chmod -R o+w laravel/storage

  6. main函数参数解析

    int argc,char *argv agrc表示参数的个数 argv储存参数 这个函数的意思是逐一输出参数 实际上,main函数也可以带参数.带参数main函数的定义格式如下:void main( ...

  7. ADO.NET增删改-------跟查不一样

    建立数据库 create database ren go use database go create table user ( code nvarchar(20) primary key,--编号 ...

  8. 插入排序和一点小感悟(c++版)

    很早之前,为了应付数据结构考试.花了一星期多看了数据结构,当时觉得也没什么难的. 过了老久,总算是招报应了,做笔试题发现其实所有理解只是在表面,实际上我并不会实现,确实是这样,学术这东西真没捷径,还是 ...

  9. 使用JavaScript

    使用JavaScript 1.在HTML中的脚本必须位于<script>和</script>之间,脚本可以被放置在HTML页面的<body>或者<head&g ...

  10. 1-NPM

    什么是NPM NPM是随同NodeJS一起安装的包管理工具,能解决NodeJS代码部署上的很多问题. 常见的使用场景有以下几种: 允许用户从NPM服务器下载别人编写的第三方包到本地使用. 允许用户从N ...