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

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
要求很清楚了。给一个数组。然后求里面出现次数最多的数字，并且求这个数字在数组里面出现的次数。

另：需要给出算法的时间复杂度。

写出算法给5分，写出算法+时间复杂度给50分。顶一下给2分。最优算法给100分。

下面是测试用的数组，一共1000个数字。

Java code

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};  

Java code

package CSDN;import java.util.ArrayList;import java.util.Collections;import java.util.HashMap;import java.util.List;import java.util.Map;public class CountArray {    public static void main(String[] args) {        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 };        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        List<Integer> list = new ArrayList<Integer>();        for (int n : nums)            // key - value (数据 - 数据出现的次数)            map.put(n, map.get(n) == null ? 1 : map.get(n) + 1);        for (int n : map.keySet()) { // 将数据出现的次数存放到ArrayList数列中            list.add(map.get(n));            System.out.println(n + "出现的次数：" + map.get(n));        }        for (int n : map.keySet()) { // 获取出现频率最高的数据            if (map.get(n) == Collections.max(list))                System.out.println("出现频率最高的是：" + n);        }    }} 
 
[解决办法]
放入hashMap  足以 然后顺搜足以啊

o（n）+顺搜一次 o（n）   事件复杂度上界 O(N) 啊
[解决办法]
学习学习，菜鸟来学习

祝楼主好运
[解决办法]
如果数组中的数字仅仅是0-9的话，可以如下实现
Java code
/*     * 返回一个数组，第一个元素放数字最多的一个那个数     * 第二个元素是出现的次数     */    public int[] getMaxTimes(int[] myArr){        int[] retArr = new int[2];        int[] myNum = new int[10];        int maxTimes = 0;        int maxNum = 0;                for(int i = 0; i < myArr.length; ++i){            ++myNum[myArr[i]];        }                //找到最大的数字和出现的次数        for(int j = 0; j < 9; ++j){            if(maxTimes < myNum[j]){                maxTimes = myNum[j];                maxNum = j;            }        }                retArr[0] = maxNum;        retArr[1] = maxTimes;                return retArr;    }
[解决办法]
探讨


如果是LZ举例的数字的话，投机取巧一下
Java code
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……

[解决办法]
探讨


如果是LZ举例的数字的话，投机取巧一下
Java code
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……

[解决办法]
原来上面有很多兄弟都是这样想的。
另：用map的时间复杂度绝对不只O(N+M)
当然如果用空间换时间的办法，定义出最小数至最大数长度的数组，分别表示最每个数的次数是可以的。。。这样的复杂度差不多有O(2N+M)
[解决办法]
如果不考虑空间的话就直接建立一个1000的数组，从0开始，直接遍历，每个数字对应数组的元素即+1，最后看哪个元素大，即出结果～～
[解决办法]
只能顶一下~
[解决办法]
阿宝的就是最经典的计数器算法，如果不知道数据范围的话：
Java code
    int[] nums = {}; // 鄙视一下lz的int nums[]写法    int max = nums[0];    int min = nums[0];    for (int i : nums) { // O(N)      if (i > max) {        max = i;      } else if (i < min) {        min = i;      }    }    System.out.printf("max: %d, min: %d%n", max, min);    int[] count = new int[max - min + 1];    int maxCnt = count[0];    int maxNum = nums[0];    for (int i : nums) { // O(N)      count[i - min]++;      if (count[i - min] > maxCnt) {        maxCnt = count[i];        maxNum = i;      }    }    System.out.printf("Number %d has most occurrences: %d%n", maxNum, maxCnt);
[解决办法]
Java code
int nums []=  { 2,4,3,6,9,2,1,9,3,7,4,1,0,1,2,2,1,5,8,8,4,1,1,7,6,1,9,6,5,5,6,9,1,8,7,9,9,1,6,3,4,4,6,5,3,1,1,5,7,7,8,0,1,9,3,5,2,4,4,4,0,8,5,3,2,1,9,6,7,9,8,2,6,4,1,7,2,0,9,1,4,6,6,5,5,6,3,5,3,3,1,6,4,4,4,0,4,4,8,3,6,3,1,0,6,2,7,1,8,4,7, 3,3,1,3,1,3,7,4,0,9, 2,1,9, 2,8,8,0,3,3,4,9,         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, ,7,7,2,3,,1,1,7,6,                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,             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       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};         //存放每个数字出现的次数        int[] count = new int[nums.length];        //遍历数组，将次数存入count         for(int i=0;i<nums.length;i++){            if(nums[i]==0){                count[0]++;            }else if(nums[i]==1){                count[1]++;            }else if(nums[i]==2){                count[2]++;            }else if(nums[i]==3){                count[3]++;            }else if(nums[i]==4){                count[4]++;            }else if(nums[i]==5){                count[5]++;            }else if(nums[i]==6){                count[6]++;            }else if(nums[i]==7){                count[7]++;            }else if(nums[i]==8){                count[8]++;            }else if(nums[i]==9){                count[9]++;            }        }        //出现最大次数        int maxNum = 0;        //存放最大数及其出现次数        String result ="";        //数组下标即对应数字，内容是对应数字出现次数        for(int i =0;i<count.length;i++){            if(count[i]>maxNum){                maxNum = count[i];                result = maxNum+"-"+i;            }        }        System.out.println("最大的数为"+result.split("-")[1]+",出现次数是"+result.split("-")[0]); 
 
[解决办法]
如果只有0－9之间的数的话，14楼的答案是很好的，14楼的答案如下。

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]);
时间复杂度O(N)，只需要遍历数组一遍即可


如果不确定数的大小，可以参考这个算法或haspmap的算法。或另做考虑。接个分吧。
[解决办法]
4楼和14楼 帅！
[解决办法]
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");
}
}


这算法有点浪费资源，呵呵。复杂度还没学。呵呵。
[解决办法]
探讨


引用:

原来上面有很多兄弟都是这样想的。
另：用map的时间复杂度绝对不只O(N+M)
当然如果用空间换时间的办法，定义出最小数至最大数长度的数组，分别表示最每个数的次数是可以的。。。这样的复杂度差不多有O(2N+M)


HashMap 查找一个 key 的复杂度是 O(1)

[解决办法]
探讨


引用:

引用:

原来上面有很多兄弟都是这样想的。
另：用map的时间复杂度绝对不只O(N+M)
当然如果用空间换时间的办法，定义出最小数至最大数长度的数组，分别表示最每个数的次数是可以的。。。这样的复杂度差不多有O(2N+M)


HashMap 查找一个 key 的复杂度是 O(1)

找到数组索引后Entry链上的遍历查找为何只有O(1)

[解决办法]
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);    } 
 
[解决办法]
就是写一个程序判断一个数组中出现次数最多的那个元素。
方法一：

import java.util.*;
public class Test1{
  public static void main(String args[]){
    String[] s = {"a","b","a","a","b","c"};
    HashMap< String, Integer> hm = new HashMap< String, Integer>();
      
    for (int i = 0; i < s.length; i++) {
       if (!hm.containsKey(s[i])) {
               hm.put(s[i], 1);
       } else  {
               hm.put(s[i], hm.get(s[i]).intValue()+1);
           }
       }
      
       System.out.println(hm);
 }
}
运行结果：

{a=3, c=1, b=2}

方法二：

import java.util.*;
public class FindMostEle {
   private static LinkedHashMap<String, Integer> map;
  
   public static LinkedHashMap<String, Integer> mostEle(String[] strArray){
       map = new LinkedHashMap<String, Integer>();
      
       String str = "";
       int count = 0;
       int result = 0;
      
       for(int i=0; i<strArray.length; i++)
           str += strArray[i];
      
       for(int i=0; i<strArray.length; i++){
           String temp = str.replaceAll(strArray[i], "");
           count = (str.length() - temp.length())/strArray[i].length();
          
           if (count > result){
               map.clear();
               map.put(strArray[i], count);
               result = count;
           }
           else if(null == map.get(strArray[i]) && count == result)
               map.put(strArray[i], count);
       }      
       return map;
   }
  
   public static void main(String args[]){
       String[] strArray = {"11", "11", "2", "2", "4", "5", "4"};
      
       LinkedHashMap<String, Integer> result = mostEle(strArray);
      
       ArrayList<Integer> c = new ArrayList<Integer>(result.values());
       Set<String> s = result.keySet();
      
       System.out.print("一共有"+ result.size() +"元素最多。它们分别是");
       System.out.print(s);
       System.out.println("，分别出现了"+ c.get(0) +"次。");
      
   }
}
结果是：

一共有3元素最多。它们分别是[11, 2, 4]，分别出现了2次。
方法三：
import java.util.*;
public class test {
   public static void main(String[] args) {
       int[] array = {1,2,2,3,4,4,5,6,7,7,8,9};
       int iMaxCount=0;
       //因为可能有多个元素的出现次数同时位居第一，
       //所以结果应该是一个集合。用一个List来装载。
       List< Integer> list=new ArrayList< Integer>();    
       for (int i=0;i< array.length;i++){
           int iCount=0;
           for (int j=0;j< array.length;j++){
                  if (array[i]==array[j])
                      iCount++;
           }
           if (iCount>iMaxCount){
               list.clear();
               list.add(array[i]);
               iMaxCount=iCount;
           }else if (iCount==iMaxCount){
               if (list.indexOf(array[i])==-1)
                   list.add(array[i]);   
           }
       }
       System.out.println("出现次数最多的元素为"+list.toString());
   }
}
运行结果：
出现次数最多的元素为[2, 4, 7]
[解决办法]
map 最简洁，但时间复杂度并非最低，而且实用map相对的占用的空间复杂度也偏高，但简单易懂，使用有上限的数组时间复杂度最低，但是空间复杂度就相对高些，而排序再计算的方式，时间复杂度和空间复杂度都蛮高的。。。