【Problem Description】
给定一副牌，每张牌上都写着一个整数。
此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：
每组都有 X 张牌。
组内所有的牌上都写着相同的整数。
仅当你可选的 X >= 2 时返回 true。
示例 1：
输入：[1,2,3,4,4,3,2,1]
输出：true
解释：可行的分组是 [1,1]，[2,2]，[3,3]，[4,4]
示例 2：
输入：[1,1,1,2,2,2,3,3]
输出：false
解释：没有满足要求的分组。
示例 3：
输入：[1]
输出：false
解释：没有满足要求的分组。
示例 4：
输入：[1,1]
输出：true
解释：可行的分组是 [1,1]
示例 5：
输入：[1,1,2,2,2,2]
输出：true
解释：可行的分组是 [1,1]，[2,2]，[2,2]
提示：
1 <= deck.length <= 10000
0 <= deck[i] < 10000
[Thinking] answer
1. The time complexity: O (N) space complexity: O (N)
- Traversal time, count the number of each value, if a value is only one direct return false;
- Enter [2, 2, 2, 2, 3, 3, 3, 3, 3, 3], in fact, is in line grouping meaning of the title, 2, 4, 3, there are six, the same 2 and 3 need to split into a set of two, i.e., whether the request twenty-two divisor number of elements have a common divisor.
public boolean hasGroupsSizeX(int[] deck) {
//计数
int[] cnt = new int[10000];
for(int x: deck){
cnt[x]++;
}
//避免虚幻内赋值
int a= cnt[0];
for(int i: cnt){
//出现单个返回错误
if(i==1){
return false;
}
if(i>1){
//循环计算最大公约数
a = gcd(a,i);
if(a ==1 )
{
return false;
}
}
}
return true;
}
/*
最大公约数
12 ➗ 9 = 1 ...... 3
9 ➗ 3 = 3 ...... 0
3返回 0跳出循环
*/
public int gcd(int x, int y){
int max = Math.max(x,y);
int min = Math.min(x,y);
while(min!= 0){
int temp =min;
min = max% min;
max = temp;
}
return max;
/*
//递归求最大公约数（伟哥提供）
private int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
}
作者：liweiwei1419
链接：https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/solution/qiu-jie-zui-da-gong-yue-shu-java-by-liweiwei1419/
2. ###### sweet Aunt streamline operations
2.1 3ms's knife
class Solution {
public boolean hasGroupsSizeX(int[] deck) {
// 计数
int[] counter = new int[10000];
for (int num: deck) {
counter[num]++;
}
// 迭代求多个数的最大公约数
int x = 0;
for(int cnt: counter) {
if (cnt > 0) {
x = gcd(x, cnt);
if (x == 1) { // 如果某步中gcd是1，直接返回false
return false;
}
}
}
return x >= 2;
}
// 辗转相除法
private int gcd (int a, int b) {
return b == 0? a: gcd(b, a % b);
}
}
作者：sweetiee
链接：https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/solution/3ms-jian-dan-java-fu-zeng-reducexie-fa-miao-dong-b/
Gcd 2.2 Iterative process can be simplified with code reduce operator
class Solution {
public boolean hasGroupsSizeX(int[] deck) {
// 计数
int[] counter = new int[10000];
for (int num: deck) {
counter[num]++;
}
// reduce求多个数的最大公约数
// （因为这里当gcd==1的时候没有提前终止，并且本题数据量太小无法用并行流，因此耗时10ms，比for循环慢点）
return Arrays.stream(counter).reduce(this::gcd).getAsInt() >= 2;
}
// 辗转相除法
private int gcd (int a, int b) {
return b == 0? a: gcd(b, a % b);
}
}
作者：sweetiee
链接：https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/solution/3ms-jian-dan-java-fu-zeng-reducexie-fa-miao-dong-b/
Wherein, reduce (this :: gcd) shall reduce ((a, b) - > gcd (a, b))
The following figure reduce ((a, b) - > a + b) is explained reduce operator:
【to sum up】
1. The greatest common divisor (Euclidean)
public int gcd(int x, int y){
int max = Math.max(x,y);
int min = Math.min(x,y);
while(min!= 0){
int temp =min;
min = max% min;
max = temp;
}
return max;
private int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
private int gcd (int a, int b) {
return b == 0? a: gcd(b, a % b);
- Count the number of times you can use a hash table, you can also use an array.
- Look at the data range of the title, if the title says, the input data contains only lowercase letters (uppercase characters) using an array of statistics is relatively easy
- The underlying hash table is an array. Using arrays do the counting tasks themselves actually prepared the hash function
- Variable name only a few can write a few words with abbreviations, under normal circumstances, all with full name.
- cnt represents the count, res represents result, ans represents the answer
