C语言如何统计二维数组中不同元素的出现次数

你现有代码中的判断条件array[j][i] == array[j][i]是恒成立的逻辑，只能统计出数组的总元素个数，无法实现按不同元素分类计数的需求。

修正后可直接运行的完整代码 #

#include <stdio.h>
int main(){
    // 定义各数字对应的计数器
    int one = 0;
    int two = 0;
    int three = 0;
    int five = 0;
    int six = 0;

    char array[3][5] = {
        {'1', '3', '3', '5', '1'},
        {'2', '1', '5', '6', '2'},
        {'6', '2', '5', '5', '2'}
    };

    // 遍历二维数组所有元素
    for(int j = 0; j < 3; j++){
        for(int i = 0; i < 5; i++){
            // 匹配到对应数字就给对应计数器+1
            switch(array[j][i]) {
                case '1': one++; break;
                case '2': two++; break;
                case '3': three++; break;
                case '5': five++; break;
                case '6': six++; break;
                default: break;
            }
        }
    }

    // 按要求格式输出结果
    printf("number one is found %d times in array\n", one);
    printf("number two is found %d times in array\n", two);
    printf("number three is found %d times in array\n", three);
    printf("number five is found %d times in array\n", five);
    printf("number six is found %d times in array\n", six);
    
    return 0;
}

代码说明 #

• 遍历二维数组的每个元素，用switch分支匹配字符类型的数字，匹配成功就给对应计数器累加，逻辑简单直观
• 如果后续数组内的数字范围扩大，可改用计数数组实现：将字符数字转为整型后作为数组下标，直接累加计数，无需写多个分支，扩展性更好

运行输出结果 #

number one is found 3 times in array
number two is found 4 times in array
number three is found 2 times in array
number five is found 4 times in array
number six is found 2 times in array

内容的提问来源于stack exchange，提问作者david