This article shares with you common algorithms familiar with in IOS interviews. Let’s take a look together below.
1. Sort the following set of data in descending order (bubble sort). “24, 17, 85, 13, 9, 54, 76, 45, 5, 63”
int main(int argc, char *argv[]) {
int array[10] = {24, 17, 85, 13, 9, 54, 76, 45, 5, 63};
int num = sizeof(array)/sizeof(int);
for(int i = 0; i < num-1; i++) {
for(int j = 0; j < num - 1 - i; j++) {
if(array[j] < array[j+1]) {
int tmp = array[j];
array[j] = array[j+1];
array[j+1] = tmp;
}
}
}
for(int i = 0; i < num; i++) {
printf("%d", array[i]);
if(i == num-1) {
printf("\n");
}
else {
printf(" ");
}
}
}
2. Sort the following set of data in ascending order (select sort). “86, 37, 56, 29, 92, 73, 15, 63, 30, 8”
void sort(int a[],int n)
{
int i, j, index;
for(i = 0; i < n - 1; i++) {
index = i;
for(j = i + 1; j < n; j++) {
if(a[index] > a[j]) {
index = j;
}
}
if(index != i) {
int temp = a[i];
a[i] = a[index];
a[index] = temp;
}
}
}
int main(int argc, const char * argv[]) {
int numArr[10] = {86, 37, 56, 29, 92, 73, 15, 63, 30, 8};
sort(numArr, 10);
for (int i = 0; i < 10; i++) {
printf("%d, ", numArr[i]);
}
printf("\n");
return 0;
}
3. Quick sorting algorithm
void sort(int *a, int left, int right) {
if(left >= right) {
return ;
}
int i = left;
int j = right;
int key = a[left];
while (i < j) {
while (i < j && key >= a[j]) {
j--;
}
a[i] = a[j];
while (i < j && key <= a[i]) {
i++;
}
a[j] = a[i];
}
a[i] = key;
sort(a, left, i-1);
sort(a, i+1, right);
}
4. Organize and sort
void merge(int sourceArr[], int tempArr[], int startIndex, int midIndex, int endIndex) {
int i = startIndex;
int j = midIndex + 1;
int k = startIndex;
while (i != midIndex + 1 && j != endIndex + 1) {
if (sourceArr[i] >= sourceArr[j]) {
tempArr[k++] = sourceArr[j++];
} else {
tempArr[k++] = sourceArr[i++];
}
}
while (i != midIndex + 1) {
tempArr[k++] = sourceArr[i++];
}
while (j != endIndex + 1) {
tempArr[k++] = sourceArr[j++];
}
for (i = startIndex; i <= endIndex; i++) {
sourceArr[i] = tempArr[i];
}
}
void sort(int souceArr[], int tempArr[], int startIndex, int endIndex) {
int midIndex;
if (startIndex < endIndex) {
midIndex = (startIndex + endIndex) / 2;
sort(souceArr, tempArr, startIndex, midIndex);
sort(souceArr, tempArr, midIndex + 1, endIndex);
merge(souceArr, tempArr, startIndex, midIndex, endIndex);
}
}
int main(int argc, const char * argv[]) {
int numArr[10] = {86, 37, 56, 29, 92, 73, 15, 63, 30, 8};
int tempArr[10];
sort(numArr, tempArr, 0, 9);
for (int i = 0; i < 10; i++) {
printf("%d, ", numArr[i]);
}
printf("\n");
return 0;
}
5. Implement binary search algorithm (programming language is not limited)
int bsearchWithoutRecursion(int array[],int low,int high,int target) {
while(low <= high) {
int mid = (low + high) / 2;
if(array[mid] > target)
high = mid - 1;
else if(array[mid] < target)
low = mid + 1;
else //findthetarget
return mid;
}
//the array does not contain the target
return -1;
}
----------------------------------------
Recursive implementation
int binary_search(const int arr[],int low,int high,int key)
{
int mid=low + (high - low) / 2;
if(low > high)
return -1;
else{
if(arr[mid] == key)
return mid;
else if(arr[mid] > key)
return binary_search(arr, low, mid-1, key);
else
return binary_search(arr, mid+1, high, key);
}
}
6. How to achieve linked list flip (reverse order of linked list)?
Idea: Every time I mention the second element to the front.
#include <>
#include <>
typedef struct NODE {
struct NODE *next;
int num;
}node;
node *createLinkList(int length) {
if (length <= 0) {
return NULL;
}
node *head,*p,*q;
int number = 1;
head = (node *)malloc(sizeof(node));
head->num = 1;
head->next = head;
p = q = head;
while (++number <= length) {
p = (node *)malloc(sizeof(node));
p->num = number;
p->next = NULL;
q->next = p;
q = p;
}
return head;
}
void printLinkList(node *head) {
if (head == NULL) {
return;
}
node *p = head;
while (p) {
printf("%d ", p->num);
p = p -> next;
}
printf("\n");
}
node *reverseFunc1(node *head) {
if (head == NULL) {
return head;
}
node *p,*q;
p = head;
q = NULL;
while (p) {
node *pNext = p -> next;
p -> next = q;
q = p;
p = pNext;
}
return q;
}
int main(int argc, const char * argv[]) {
node *head = createLinkList(7);
if (head) {
printLinkList(head);
node *reHead = reverseFunc1(head);
printLinkList(reHead);
free(reHead);
}
free(head);
return 0;
}
7. Implement the reverse order output of a string "how are you" (the programming language is not limited). If the given string is "hello world", the output result should be "world hello".
int spliterFunc(char *p) {
char c[100][100];
int i = 0;
int j = 0;
while (*p != '\0') {
if (*p == ' ') {
i++;
j = 0;
} else {
c[i][j] = *p;
j++;
}
p++;
}
for (int k = i; k >= 0; k--) {
printf("%s", c[k]);
if (k > 0) {
printf(" ");
} else {
printf("\n");
}
}
return 0;
}
8. Given a string, output the position of the character that appears only once and is the first in this string? For example, "abaccddeeef", the character is b, and the output should be 2.
char *strOutPut(char *);
int compareDifferentChar(char, char *);
int main(int argc, const char * argv[]) {
char *inputStr = "abaccddeeef";
char *outputStr = strOutPut(inputStr);
printf("%c \n", *outputStr);
return 0;
}
char *strOutPut(char *s) {
char str[100];
char *p = s;
int index = 0;
while (*s != '\0') {
if (compareDifferentChar(*s, p) == 1) {
str[index] = *s;
index++;
}
s++;
}
return &str;
}
int compareDifferentChar(char c, char *s) {
int i = 0;
while (*s != '\0' && i<= 1) {
if (*s == c) {
i++;
}
s++;
}
if (i == 1) {
return 1;
} else {
return 0;
}
}
9. The preorder traversal of the binary tree is FBACDEGH, and the mid-order traversal is: ABDCEFGH. Please write the result of the postorder traversal of this binary tree.
ADECBHGF
Preorder + middle order traversal to restore the binary tree: preorder traversal is: ABDEGCFH In-order traversal is: DBGEACHF
First, we get the first one from the predecessor order, which is the root of the binary tree. Return to the middle order, we can divide it into three parts:
The middle sequence DBGE of the left subtree, root A, and the middle sequence CHF of the right subtree
Then return the sequence of the left subtree to the precedent order to get B as the root, so return to the middle order of the left subtree and divide the left subtree into three parts:
The left subtree D of the left subtree, the root B of the left subtree, the right subtree GE of the left subtree
Similarly, the root of the right subtree can be obtained as C
Similarly, the right subtree is divided into root C, the right subtree HF of the right subtree, note that its left subtree is empty
If there is only one leaf, no more, and the GE and HF just now operate in this way again, you can restore the binary tree.
10. Print prime numbers between 2-100.
int main(int argc, const char * argv[]) {
for (int i = 2; i < 100; i++) {
int r = isPrime(i);
if (r == 1) {
printf("%ld ", i);
}
}
return 0;
}
int isPrime(int n)
{
int i, s;
for(i = 2; i <= sqrt(n); i++)
if(n % i == 0) return 0;
return 1;
}
11. Find the greatest common divisor of two integers.
int gcd(int a, int b) {
int temp = 0;
if (a < b) {
temp = a;
a = b;
b = temp;
}
while (b != 0) {
temp = a % b;
a = b;
b = temp;
}
return a;
}
Summarize
The above are the common algorithm questions and answers that you may encounter in IOS interviews. I hope this article will be of some help to your interview. If you have any questions, you can leave a message to communicate.