IOS面试大全之常见算法

这篇文字给大家分享了IOS面试中熟悉常见的算法,下面来一起看看吧。

1、 对以下一组数据进行降序排序(冒泡排序)。“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、 对以下一组数据进行升序排序(选择排序)。“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、 快速排序算法

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、 归并排序

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、 实现二分查找算法(编程语言不限)

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;

}

----------------------------------------

递归实现

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、 如何实现链表翻转(链表逆序)?

思路:每次把第二个元素提到最前面来。

#include <stdio.h>

#include <stdlib.h>


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、 实现一个字符串“how are you”的逆序输出(编程语言不限)。如给定字符串为“hello world”,输出结果应当为“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、 给定一个字符串,输出本字符串中只出现一次并且最靠前的那个字符的位置?如“abaccddeeef”,字符是b,输出应该是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、 二叉树的先序遍历为FBACDEGH,中序遍历为:ABDCEFGH,请写出这个二叉树的后序遍历结果。

ADECBHGF

先序+中序遍历还原二叉树:先序遍历是:ABDEGCFH 中序遍历是:DBGEACHF

首先从先序得到第一个为A,就是二叉树的根,回到中序,可以将其分为三部分:

左子树的中序序列DBGE,根A,右子树的中序序列CHF

接着将左子树的序列回到先序可以得到B为根,这样回到左子树的中序再次将左子树分割为三部分:

左子树的左子树D,左子树的根B,左子树的右子树GE

同样地,可以得到右子树的根为C

类似地将右子树分割为根C,右子树的右子树HF,注意其左子树为空

如果只有一个就是叶子不用再进行了,刚才的GE和HF再次这样运作,就可以将二叉树还原了。

10、 打印2-100之间的素数。

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、 求两个整数的最大公约数。

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;

}

总结

以上就是为大家整理的在IOS面试中可能会遇到的常见算法问题和答案,希望这篇文章对大家的面试能有一定的帮助,如果有疑问大家可以留言交流。