iOS面试题系列之常见算法

电玩女神 2023-10-17 13:23 31阅读 0赞

原文链接[点击查看][Link 1]

iOS面试中熟悉常见算法

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;
    }

<table> 
 <tbody> 
  <tr> 
   <td style="width:645.867px;"> <pre>&nbsp;</pre> 
    <div>
      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”。 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

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;
    }

<table> 
 <tbody> 
  <tr> 
   <td style="width:645.867px;"> <pre>&nbsp;</pre> 
    <div>
      11、 求两个整数的最大公约数。 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

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;
    }

[Link 1]: https://www.yangshebing.com/2016/04/24/iosmian-shi-ti-xi-lie-zhi-chang-jian-suan-fa/