騰訊筆試題

關于騰訊筆試題

部分IT公司筆試算法題

1、將一整數逆序后放入一數組中(要求遞歸實現)

void convert(int *result, int n) {

if(n>=10)

convert(result+1, n/10);

*result = n%10;

}

int main(int argc, char* argv[]) {

int n = 123456789, result[20] = { };

convert(result, n);

printf("%d:", n);

for (int i = 0; i < 9; i++)

printf("%d", result);

}

2、求高于平均分的學生學號及成績(學號和成績人工輸入)

double find(int total, int n) {

int number, score, average;

scanf("%d", &number);

if(number != 0) {

scanf("%d", &score);

average = find(total+score, n+1);

if(score >= average)

printf("%d:%d ", number, score);

return average;

} else {

printf("Average=%d ", total/n);

return total/n;

}

}

int main(int argc, char* argv[]) {

find(0, 0);

}

3、遞歸實現回文判斷(如：abcdedbca就是回文，判斷一個面試者對遞歸理解的簡單程序)

int find(char *str, int n) {

if(n<=1) return 1;

else if(str[0]==str[n-1]) return find(str+1, n-2);

else return 0;

}

int main(int argc, char* argv[]) {

char *str = "abcdedcba";

printf("%s: %s ", str, find(str, strlen(str)) ? "Yes" : "No");

}

4、組合問題(從M個不同字符中任取N個字符的所有組合)

void find(char *source, char *result, int n) {

if(n==1) {

while(*source)

printf("%s%c ", result, *source++);

} else {

int i, j;

for(i=0; source != 0; i++);

for(j=0; result[j] != 0; j++);

for(; i>=n; i--) {

result[j] = *source++;

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