iOS 算法~面试题

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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、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

#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;

}

4、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;

}

5、打印2-100之间的素数。

[cpp] view plain copy

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;

}

6、求两个整数的最大公约数。

[cpp] view plain copy

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;

}

7、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;

}

8、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。

9、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;

}

10、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”。

[cpp] view plain copy

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;

}

11、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。

[cpp] view plain copy

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;

}

}

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