推 applea123:感謝提供 讓我研究下 10/21 17:46
→ tropical72:樓上 zainan 廣告文太扯了唷.. 10/21 21:39
※ 引述《applea123 (小刀)》之銘言：
: 最近在寫眾數問題
: 主要是考慮到眾數>2時
: ex:1 1 1 1 1 2 2 2 2 3 3 3 4 4
: 眾數有1 2
: 次數4
: 我自己是有寫出來 但是想問問大家有沒更快的"想法"
: 我的步驟:
: 1.排序 並丟入a[]
: 2.用一個陣列b[]紀錄各數出現次數並使用另一個對應陣列c[]來記錄此數字
: 3.陣列b排列並且c一起做交換
: 4.判斷b[] if(b[i]=b[i+1])
: 5.印出b[] 與c[]
我的作法沒很快,很基本:
一直線掃描,看過的數字把頻次加一,沒看過的數字填入表中.
(見 find函數)
struct Freq {
int n;
int times;
struct Freq *next;
};
struct Freq* find(int list[], int len) {
struct Freq* lookup(struct Freq *list, int n);
int i;
struct Freq *result = NULL;
for (i=0; i<len; i++) {
struct Freq *p, *q;
p = lookup(result, list[i]);
if (p == NULL) {
q = (struct Freq*)malloc(sizeof(struct Freq));
q->n = list[i];
q->times = 1;
q->next = result;
result = q;
} else {
p->times++;
}
}
return result;
}
struct Freq* lookup(struct Freq *list, int n) {
struct Freq *p = list;
while (p != NULL) {
if (p->n == n) return p;
p = p->next;
}
return p;
}
如此可以很簡單找到數頻表:
int main()
{
struct Freq* find(int list[], int len);
void layout(struct Freq *list);
int list[14] = {1,1,3,1,4,2,4,2,2,3,3,1,2,1};
struct Freq *flist = find(list, 14);
layout(flist);
system("pause");
return 0;
}
void layout(struct Freq *list) {
struct Freq *p = list;
while (p!= NULL) {
printf("%2d%2d\n", p->n, p->times);
p = p->next;
}
}
輸出為:
2 4
4 2
3 3
1 5
Press any key to continue . . .
要不要排序隨便你. 如果只要找眾數群:
int main()
{
struct Freq* find(int list[], int len);
void layout(struct Freq *list, int d);
int list[14] = {1,1,3,1,4,2,4,2,2,3,3,1,2,1};
struct Freq *flist = find(list, 14);
layout(flist, 2);
system("pause");
return 0;
}
void layout(struct Freq *list, int d) {
struct Freq *p = list;
int max = 0;
printf("Modes: ");
while (p!= NULL) {
if (p->times > d)
printf("%2d", p->n);
if (p->times > max)
max = p->times;
p = p->next;
}
printf("\n");
printf("Maximal frequency: %d\n", max);
}
輸出為:
Modes: 2 3 1
Maximal frequency: 5
Press any key to continue . . .
完成.
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