Question

我可以在数组中找到重复的值，但是当我有两组以上的重复值时程序会卡住。

例如，如果我给出输入：

public function index()
{
    $this->load->model('sorts');
    $data['products'] = $this->sorts->sort_products1();

    $this->load->view('cart/products', $data);
}

（注意第一个数字[10]表示数组中有多少个数字，并且假设所有数字都按升序排列）

我得到

的正确分数输出

10 1 2 3 4 5 10 10 10 10 10

（runCount（5）* runNumber（10）= 50）

如果我在数组中有两个以上的重复元素，例如 -

Run {10,10,10,10,10} scoring 50.

分数都是混乱的

10 5 5 5 5 5 10 10 10 10 10

由于{5,5,5,5,5}的得分应低于{10,10,10,10,10}，理想情况下，该程序会打印得分最高的值。

 Run {5,10,10,10,10,5,5,5,5} scoring 45.

如何制作它只使用最高值？我无法让break函数正常工作。

如果您有任何想法，请告诉我？我的代码如下。

我只能使用数组，循环和if / else

Run {10,10,10,10,10} scoring 50.

Answer 1

像这样：

#include <stdio.h>
#include <limits.h>

int main(void){
    int n;
    scanf("%d", &n);

    int numbers[n];
    for(int i = 0; i < n; ++i) {
        scanf("%d", &numbers[i]);
    }

    int hi_score = INT_MIN, dupCount = 0 , theNumber;
    for(int i = 0; i < n; ++i){
        int counter = 0, score;
        for(int j = i + 1; j < n && numbers[i] == numbers[j]; ++j){
            ++counter;
        }
        if(counter > 0){
            score = (counter + 1) * numbers[i];
            if(hi_score <= score){//= : when 4 4 4 4 4 5 5 5 5, 4 < 5
                theNumber = numbers[i];
                dupCount = counter + 1;
                hi_score = score;
            }
            i += counter;
        }
    }
    if(dupCount > 0){
        int new_array[dupCount];

        printf("Run {");
        for(int i = 0; i < dupCount; ++i){
            if(i)
                putchar(',');
            printf("%d", new_array[i] = theNumber);
        }
        printf("} scoring %d.\n", hi_score);
    }

    return 0;
}

Answer 2

或尝试这样：

#include<stdio.h>
#include<stdlib.h>


void generateScore(int*, int , int* , int* );

int main(){
    int number = 0;
    int times = 0;
    int array[10] = {10, 10, 10, 10, 10, 4, 4, 10, 10, 10};

    generateScore(array, 10, &number, &times);
    printf("number: %d \t times:%d", number, times);

    return (0);
}


void generateScore(int* array, int size, int* number, int* times) {
    *number = array[0];
    *times = 1;
    int i;
    for(i = 1 ; i < size && array[i] == *number; i++) {
        *times = *times + 1 ;
    }
    if(size <= i) {
        return;
    }

    int number2 = array[i];
    int times2 = 1;
    i++;
    for(; i < size; i++){
        if(array[i] == number2){
            times2 = times2 + 1;
        } else {
            if((*times * (*number)) < (number2 * times2)) {
                *times = times2;
                *number = number2;
            }
            number2 = array[i];
            times2 = 1;
        }
    }
    if((*times * (*number)) < (number2 * times2)) {
        *times = times2;
        *number = number2;
    }
}

Answer 3

int array[10] = {5 5 5 5 5 10 10 10 10 10};

generateScore(array, 10, &number, &times);
void generateScore(....){
  int sum=0;
  int preSum=0;
  int index=-1;
  int k;
  for(i=0;i<size;i++)
  {
   k=i;
   sum=0;
   while(((k+1)<size)&&(array[k]==array[k+1]))
    {
     sum=sum+array[k];
     k=k+1;
    }
   if(sum>presum){
    index=k-1;
    presum=sum;
   }
   i=k;
  }
var num = presum/array[index];
//now your sum in presum which is present num times in your array whose last index in array is index
}

了解更多see here

如何在数组中找到匹配的值并将它们存储在新数组中？

3 个答案: