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1、對以下一組數(shù)據(jù)進行降序排序（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、對以下一組數(shù)據(jù)進行升序排序（選擇排序）。“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、如何實現(xiàn)鏈表翻轉(zhuǎn)（鏈表逆序）？
思路：每次把第二個元素提到最前面來。
#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、給定一個字符串，輸出本字符串中只出現(xiàn)一次并且最靠前的那個字符的位置？如“abaccddeeef”,字符是b,輸出應(yīng)該是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之間的素數(shù)。
[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、求兩個整數(shù)的最大公約數(shù)。
[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、歸并排序
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、二叉樹的先序遍歷為FBACDEGH,中序遍歷為：ABDCEFGH,請寫出這個二叉樹的后序遍歷結(jié)果。
ADECBHGF
先序+中序遍歷還原二叉樹：先序遍歷是：ABDEGCFH中序遍歷是：DBGEACHF
首先從先序得到第一個為A，就是二叉樹的根，回到中序，可以將其分為三部分：
左子樹的中序序列DBGE，根A，右子樹的中序序列CHF
接著將左子樹的序列回到先序可以得到B為根，這樣回到左子樹的中序再次將左子樹分割為三部分：
左子樹的左子樹D，左子樹的根B，左子樹的右子樹GE
同樣地，可以得到右子樹的根為C
類似地將右子樹分割為根C，右子樹的右子樹HF，注意其左子樹為空
如果只有一個就是葉子不用再進行了，剛才的GE和HF再次這樣運作，就可以將二叉樹還原了。
9、打印2-100之間的素數(shù)。
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、實現(xiàn)一個字符串“how are you”的逆序輸出（編程語言不限）。如給定字符串為“hello world”,輸出結(jié)果應(yīng)當為“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、給定一個字符串，輸出本字符串中只出現(xiàn)一次并且最靠前的那個字符的位置？如“abaccddeeef”,字符是b,輸出應(yīng)該是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;
}
}
