输出array中出现次数最多的数

Question

用类似这种的方法 不用Map

int[] a = new int[10]{1,2,3,4,5,6,7,7,7,7};

public int getPopularElement(int[] a)
{
int count = 1, tempCount;
int popular = a[0];
int temp = 0;
for (int i = 0; i < (a.length - 1); i++)
{
temp = a[i];
tempCount = 0;
for (int j = 1; j < a.length; j++)
{
if (temp == a[j])
tempCount++;
}
if (tempCount > count)
{
popular = temp;
count = tempCount;
}
}
return popular;
}

我想让this.method = 那个出现次数最多的数字 如果所有数都只出现一次的话就this.method = -1
或者出现次数相同的话也是this.method = -1 。
我照着上面的方法写的没法判断每个数只出现一次的情况 还有出现次数相同的情况
求帮忙看看怎么写

public Long getMode() {

    Vector vector = cloned();
    long [] num = vector.getElements();


    // 要写的地方


    return this.method;
    }

Answer 1

 public static int getPopularElement(int[] a)
{
    if(a.length < 2){
        return a[0];
    }
    int[] t = new int[a.length];//记录每一个元素出现的次数
    int popular = -1;
    for (int i = 0; i < a.length; i++)
    {
        int temp = a[i];
        int count = 1;
        for (int j = 0; j < a.length; j++)
        {
            if(j == i){ 
                continue;
            }
            if (temp == a[j])
                count++;
        }
        t[i] = count;
    }
    //System.out.println(Arrays.toString(t));
    int first = t[0] > t[1] ? t[0] : t[1];
    int pos = t[0] > t[1] ? 0 : 1;
    int second = t[0] < t[1] ? t[0] : t[1];
    for(int i = 2; i < t.length; i++){
        if(t[i] >= first){
            first = t[i];
            pos = i;
            continue;
        }
        if(t[i] >= second){
            second = t[i];
        }
    }
    //System.out.println(first + "," + second);
    if(first == second ){
        return -1;
    }else{
        return a[pos];
    }
}