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	//findthetargetreturn mid;}//the array does not contain the targetreturn -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);elsereturn 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;}

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iOS面试题系列之常见算法