# 300分！给一个数组 求里面出现次数最多的元素和其次数 另：需要加上时间复杂度，否则只给5分。最优算法给100分,该怎么处理(3)

www.MyException.Cn  网友分享于：2013-01-16  浏览：81次

------解决方案--------------------

int[] cnt = {0,0,0,0,0,0,0,0,0,0}; //分别记录数字0,1,2,3,4,5,6,7,8,9的次数
int[] max = {0,0}; //记录出现最多的数字以及出现次数，max[0]次数，max[1]数字
for (int n : num) { //遍历数组
cnt[n]++; //该数字对应的计数器累加
if (cnt[n] > max[0]) { //比较计数器和当前最大次数
max[0] = cnt[n];
max[1] = n;
}
}
System.out.printf("最大数字：%d，出现次数：%d\n", max[1], max[0]);

------解决方案--------------------
4楼和14楼 帅！
------解决方案--------------------
public class MaxArray {

private int a []={1,2,1,1,2};
public static void main(String[] args)
{
MaxArray ob = new MaxArray();
ob.maxArray();
}
public void maxArray()
{
int b = 0 ;
int tem = 0;
int result = 0;
for(int i = 0;i<a.length;i++)
{
int count = 1;
for(int j = i+1;j<a.length;j++)
{
if((a[i]-a[j])==0)
{
count++;
}
}
if(tem<count)
{
tem = count;
b = a[i];
}
}
System.out.print("出现频率最多的数是："+b+"，一共出现了"+tem+"次");
}
}

------解决方案--------------------

HashMap 查找一个 key 的复杂度是 O(1)

------解决方案--------------------

HashMap 查找一个 key 的复杂度是 O(1)

------解决方案--------------------
Java code
```    public static void main(String[] args) {
int i,j,h,max=0,num=0;
int nums []=  {2,4,3,6,9,2,1,9,3,7,4,1,0,1,2,2,1,5,8,8,4,1,1,7,6,1,9,6,5,5,6,9,1,8,7,9,9,1,6,3,4,4,6,5,3,1,1,5,7,7,8,0,1,9,3,5,2,4,4,4,0,8,5,3,2,1,9,6,7,9,8,2,6,4,1,7,2,0,9,1,4,6,6,5,5,6,3,5,3,3,8,3,6,3,1,0,6,2,7,1,8,4,7,3,3,1,3,1,3,7,4,0,9,7,9,4,6,0,2,1,9,2,8,8,0,1,6,4,4,4,0,4,4,3,3,4,9,3,0,3,5,6,3,2,4,7,5,1,3,7,4,4,4,8,5,7,9,8,7,1,3,1,0,7,8,4,6,9,5,2,8,5,2,7,3,8,4,4,6,3,5,9,2,8,9,8,7,1,6,5,1,4,4,1,5,8,7,5,9,0,7,4,6,5,3,6,1,2,9,6,2,1,1,8,4,1,2,6,6,0,0,4,9,0,8,9,5,7,7,2,3,8,4,2,7,8,7,5,3,4,4,6,9,5,0,4,8,7,1,6,3,1,9,7,4,9,0,4,0,3,5,8,0,8,6,7,8,1,0,2,0,8,3,0,1,1,7,6,6,2,2,8,1,7,8,2,5,2,5,8,4,0,0,2,4,2,8,3,4,2,5,6,4,4,8,8,7,2,3,8,0,0,6,4,2,1,6,7,1,5,3,7,1,2,3,4,5,1,8,1,5,1,3,1,8,5,1,0,1,1,8,4,6,2,6,8,0,2,7,4,3,0,8,5,0,1,2,8,1,2,9,0,4,8,5,5,8,1,5,8,9,5,2,9,8,2,2,0,2,6,1,9,0,5,8,4,6,9,1,6,8,0,7,0,3,5,5,9,5,9,9,6,1,9,2,8,2,7,6,8,0,8,7,1,4,2,2,7,6,1,7,0,9,7,6,4,4,3,3,5,7,7,8,9,6,9,6,3,7,6,3,2,5,2,5,8,2,7,1,0,7,7,5,6,8,9,2,3,4,9,6,1,3,9,1,9,0,5,6,7,0,7,6,0,5,3,7,4,5,0,2,1,1,8,2,4,8,4,5,4,8,2,2,6,1,7,0,9,3,1,5,4,0,4,2,9,4,5,4,7,7,0,2,1,7,5,3,5,6,0,7,3,6,5,9,5,4,4,1,6,5,6,1,1,4,2,1,9,7,4,6,3,6,9,1,3,4,0,1,8,1,5,0,3,4,9,3,1,7,8,2,2,2,2,5,1,4,0,2,2,2,1,2,9,9,8,2,9,6,0,4,9,5,2,1,3,7,0,2,5,8,5,0,2,7,0,2,6,2,4,7,6,1,2,3,5,7,5,6,2,1,5,0,2,1,2,3,4,7,8,3,8,0,8,2,5,5,8,0,1,3,3,7,5,9,8,8,1,5,0,5,4,1,9,2,1,0,9,1,0,6,7,8,4,9,7,9,0,0,8,2,5,1,7,0,7,5,8,5,9,6,6,6,3,1,5,0,8,3,0,3,4,8,4,3,3,9,3,0,7,8,9,3,4,5,6,0,9,7,2,1,1,8,4,6,0,3,4,7,4,6,7,9,3,0,3,9,6,9,2,2,2,5,5,9,0,0,6,2,4,1,1,7,6,3,9,5,6,6,6,8,8,1,2,0,4,3,0,9,4,5,4,9,1,7,7,2,1,1,0,0,0,9,3,3,1,9,3,3,7,3,7,1,2,2,6,0,4,8,9,0,7,8,3,8,3,4,1,2,8,1,6,0,1,2,1,9,2,8,3,2,5,8,5,9,3,2,1,6,9,7,6,5,3,0,1,4,4,2,8,4,4,2,0,9,0,7,7,7,6,4,0,6,9,2,4,7,2,3,9,9,8,3,8,3,8,1,3,6,9,8,5,8,1,4,2,0,2,9,4,2,7,6,1,9,6,3,0,4,1,8,1,3,4,5,2,0,5,5,3,8,0,9,0,1,3,4,8,1,5,0,9,6,5,0,7,2,8,2,3,2,9,9,2,4,7,0,8,6,4,0,8,6,8,3,6,6,7,5,3,0,1,3,7,1,6,3,8,4,5,4,1,2,9,0,5,7,1,7,1,2,9,3,8,1,2,7,5,3,8,7,1,7,3,0,8,2,9,6,8,6,0,9,9,5,4,3,8,1,2,2,2,2,3,9,8,5,5,5,5,2,9,8,8,1,3,0,1,1,9,3,7,7,7,2,2,3,6,3,7};
int []count={0,0,0,0,0,0,0,0,0,0};//初始化数组记录每个数出现的次数,比如count[0]的值就是0出现的次数
//遍历nums[],记录每个数出现的次数
for(i=0;i<nums.length;i++)
{
h=nums[i];
count[h]=count[h]+1;

}
//获取count[]中最大的值，就是出现的次数，下标就是原来的数值
for(j=0;j<count.length;j++)
{
if(num<count[j])
{
num=count[j];
max=j;
}
}
System.out.println("最大数"+max);
System.out.println("出现次数"+num);
}
```