Given an array of size n, find the majority of elements in it: elements that occur more than n/2
import java.util.HashMap;
import java.util.Map;
import java.util.Set;
public class ArrayHomeWork {
/*给定一个大小为n的数组，找到其中的多数元素：出现次数大于n/2的元素
* 哈希表方法，映射存储出现元素以及该元素出现的次数*/
public int manyNum(int[] num){
//JDK的哈希表用法
HashMap<Integer,Integer> hashMap = new HashMap<>();
//遍历数组的过程中将不再重复的元素及其出现的次数存储到应设中
for (int i = 0; i < num.length; i++) {
//若此时映射中存在元素
if (hashMap.containsKey(num[i])){
//判断此时数据集中是否包含元素key -> num[i]
//此时只需要将出现次数+1
int value = hashMap.get(num[i]);//从映射中取出该key值对应的value值
hashMap.put(num[i],value + 1);
}else{
//若此时映射中还没有元素
//哈希表的存储方法 hashMap.put(不重复的元素，value）
hashMap.put(num[i],1);//向映射中存储一对元素
}
}
//在数据集中找到出现次数 》 n/2 的元素
//哈希表遍历
Set<Map.Entry<Integer,Integer>> set = hashMap.entrySet();
for (Map.Entry<Integer,Integer> entry : set){
//取出每个元素出现的次数-value值
if (entry.getValue() > num.length / 2){
return entry.getKey();
}
}
return -1;
}
public static void main(String[] args) {
ArrayHomeWork homeWork = new ArrayHomeWork();
int[] num1 = {
2 , 2 , 4 , 2};
int[] num2 = {
5 , 7 , 9 , 1};
System.out.println(homeWork.manyNum(num1));
System.out.println(homeWork.manyNum(num2));
}
}
// Determine whether there are three consecutive elements in the array that are odd numbers //
public boolean consecutiveOdds(int[] num){
int count = 0;//记录当前数组中出现连续奇数的个数
for (int i : num){
//for-each数组：不修改原数组
if (i % 2 != 0 ){
count++;
}else{
count = 0;//i是偶数，重置count变量
}
if (count == 3){
return true;
}
}
return false;
}*/
/*public static void main(String[] args) {
ArrayHomeWork homeWUork = new ArrayHomeWork();
int[] num1 = {2 , 6 , 4 , 1};
int[] num2 = {5 , 7 , 9 , 1};
System.out.println(homeWork.consecutiveOdds(num1));
System.out.println(homeWork.consecutiveOdds(num2));
}
}