• 按序输出整数数组中所有出现频率最高的数字


    蛇年过后第一次开工。这次来讨论一个简单的Amazon笔试题,我先抛出来自己的解法,欢迎大家回复更好的算法,我会更新到帖子里。大家一起学习!

    题目来自GeeksForGeeks。那是一位牛牛经过10轮笔试+面试成功应聘Amazon后回馈大众的回忆录,还有好几个题目很有意思呢,大家去看看。也许是很久之前的题了,并不见得是Amazon发明滴。

    话说今天这个题目大意如下:
    Find maximum frequent numbers in an array. If there are more than one numbers with maximum frequency, they display all numbers in ascending order. Ascending order is important.

    看起来确实很简单,如果数字大小有限,干脆开辟一个新数组a,a的每个元素i代表数字i的出现次数,遍历整数数组更新a即可。如果数字非常大又怎么办呢?思索了半天只想出来下面两种差不多的半吊子算法,既不快也不漂亮:

    算法一:
    1、对数组快速排序
    2、遍历一遍数组,记录每个数字出现次数到结构体数组frequency,顺便记录最大次数。frequency每个元素记录[integer, frequency]。
    3、遍历第2步中frequency,输出所有次数最大的数字
    时间复杂度O(nlgn),空间复杂度O(n)

    算法二:
    1、开辟一个新的数组frequency记录数字出现次数
    2、对数组进行插入排序,查找步骤中如果需要插入,则记录该数字出现次数为1;如果遇到相等的情况,就不插入,而是将相等的数字出现次数加1;插入同时更新frequency数组,要保证元素和已排序子数组各个元素一一对应
    3、遍历frequency数组,找到最大次数
    4、再次遍历frequency数组,输出所有次数最大的数字
    时间复杂度O(n^n),空间复杂度O(n)

    下面是算法二的实现。

    View Code
    /*
     * most-frequent-number.c
     *
     * by Wang Guibao
     *
     * Amazon tech interview written test problem.
     * http://www.geeksforgeeks.org/amazon-interview-set-21/
     *
     * Problem:
     * Find maximum frequent numbers in an array. If there are more numbers with
     * maximum frequency, they display all numbers in ascending order. Ascending
     * order is important.
     */
    #include <stdio.h>
    
    #define MAX_ELE 32
    
    int elements[MAX_ELE + 1];
    int frequency[MAX_ELE + 1];
    /*
     * Find most frequently appeared number in an array. If there're more than one
     * such numbers, output them in ascending order
     * @param elements: integer array, integers are elements[1...n]
     * @param n       : number of integers in array
     */
    void most_frequent_number(int *elements, int n)
    {
        int sorted_begin;
        int sorted_end;
        int begin;
        int end;
        int middle;
        int i;
        int j;
        int max_freq = 0;
    
        sorted_begin = sorted_end = 1;
        frequency[sorted_begin] = 1;
    
        /* Insert sort the array */
        for (i = 2; i <= n; i++) {
            begin = sorted_begin;
            end = sorted_end;
    
            while (begin <= end) {
                middle = (begin + end) / 2;
                if (elements[middle] > elements[i]) {
                    end = middle - 1;
                }
                else if (elements[middle] < elements[i]) {
                    begin = middle + 1;
                }
                else if (elements[middle] == elements[i]) {
                    frequency[middle]++;
                    break;
                }
            }
    
            if (end < begin) {
                for (j = sorted_end; j >= begin; j--) {
                    elements[j + 1] = elements[j];
                    frequency[j + 1] = frequency[j];
                }
                elements[begin] = elements[i];
                frequency[begin] = 1;
    
                sorted_end++;
            }
        }
    
        /* Traverse the sorted subarray */
        for (j = 1; j <= sorted_end; j++) {
            if (max_freq < frequency[j]) {
                max_freq = frequency[j];
            }
        }
    
        /* Output the integer(s) with maximum frequency */
        for (j = 1; j <= sorted_end; j++) {
            if (frequency[j] == max_freq) {
                printf(" %d", elements[j]);
            }
        }
        printf("\n");
    
        return;
    }
    
    int main()
    {
        int i = 1;
        int n;
    
        printf("Number of integers to process: ");
        scanf("%d", &n);
    
        printf("Input %d integers: ", n);
        for (i = 1; i <= n; i++) {
            scanf("%d", &elements[i]);
        }
    
        most_frequent_number(elements, n);
    
        return 0;
    }

    不知道这个题目时间、空间复杂度能否进一步降低?欢迎大家一起讨论哈。

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  • 原文地址:https://www.cnblogs.com/wangguibao/p/most_frequent_integers_in_an_array.html
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