# 计算字符串中数字的出现次数

``#include  #include  int main() { int numbers = {1, 4, 5, 5, 5, 6, 6, 3, 2, 1}; int count = 0; for(int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if (numbers[i] == numbers[j]) { count++; } } printf("Number %d has occured %d many times \n", numbers[i], count); count = 0; } }` `

` `Number: 1 Occurence: 2 Number: 4 Occurence: 1 Number: 5 Occurence: 3 Number: 5 Occurence: 3 Number: 5 Occurence: 3 Number: 6 Occurence: 2 Number: 6 Occurence: 2 Number: 3 Occurence: 1 Number: 2 Occurence: 1 Number: 1 Occurence: 2` `

` `#include  #include  int main() { int numbers = {1, 4, 5, 5, 5, 6, 6, 3, 2, 1}; int count = 0; for(int i = 0; i < 10; i++) { //i = current digit for (int j = 0; j < 10; j++) { //j = index in array if (i == numbers[j]) { count++; } } printf("Number %d has occured %d times \n", i, count); count = 0; } }` `

` `Number 0 has occured 0 times Number 1 has occured 2 times Number 2 has occured 1 times Number 3 has occured 1 times Number 4 has occured 1 times Number 5 has occured 3 times Number 6 has occured 2 times Number 7 has occured 0 times Number 8 has occured 0 times Number 9 has occured 0 times` `

` `#include  #include  int main() { int inputNumbers [] = {1, 4, 5, 5, 5, 6, 6, 3, 2, 1}; int resultCount [] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; int countNumbers = sizeof(inputNumbers) / sizeof(inputNumbers); for(int i = 0; i < countNumbers; i++) { resultCount[inputNumbers[i]]++; } for(int i = 0; i < countNumbers; i++) { printf("Number %d has occured %d times \n", i, resultCount[i]); } }` `