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）
时间： 2024-10-06 05:45:02