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This article mainly introduces how to sort and find C++, the article is very detailed, has a certain reference value, interested friends must read it!

1. Sort the array by bubbling (ascending order)

# include # include

Void sort (int a [], int n) {int I, j, t; for (I = 0; I

< n-1 ; i++) { for (j = 0; j < n-1; j++) { if (a[j] >

A [juni1]) {t = a [j]; a [j] = a [juni1]; a [juni1] = t;}

Int main () {int a [5], I; for (I = 0; I

< 5; i++) { printf("input a[%d]=", i); scanf("%d", &a[i]); } sort(a, 5); for (i = 0; i < 5; i++) { printf("%4d", a[i]); } system("pause"); return 0;} 2、用插入法排序将输入到数组中的元素进行排序 #include #include void fun(int a[], int n){ int k, j, t; for (j = 1; j < n; j++) { t = a[j]; k = j - 1; while (k >

= 0 & t > a [k]) {a [k + 1] = a [k]; kmuri;} a [k + 1] = t;}}

Int main () {int a [5], I; for (I = 0; I

< 5; i++) { printf("input a[%d]=", i); scanf("%d", &a[i]); } fun(a, 5); for (i = 0; i < 5; i++) { printf("%4d", a[i]); } system("pause"); return 0;} 3、用选择法将输入的10个数进行排序 #include #include int main(){ int i, j, min, tem, a[10]; for (i = 0; i < 10; i++) { printf("input a[%d]=", i); scanf("%d", &a[i]); } for (i = 0; i < 10; i++) { printf("%d", a[i]); } for (i = 0; i < 10 - 1; i++) { min = i; for (j = i + 1; j < 10; j++) { if (a[min] >

A [j]) {min = j;} tem = a [I]; a [I] = a [min]; a [min] = tem;} printf ("\ n"); for (I = 0; I < 10; iTunes +) {printf ("% d", a [I]);} system ("pause"); return 0;}

4. Using the half-and-half search method, you will check whether the entered number is in the ascending array of 10 elements, and if so, which number it is.

# include # include

Int main () {int a [10] = {12 int a [10] = {12 Peregra [10] = {12 Peregrine 32 [10] = {12 Peregrine 76pyrorn 85pyrorn]; int num, bott, top, mid; printf ("input a number:"); scanf ("% d", & num); bott = 0; top = 9; while (bott top) {printf ("no\ n");} system ("pause"); return 0;} above are all the contents of this article entitled "how to sort and find C++". Thank you for reading! Hope to share the content to help you, more related knowledge, welcome to follow the industry information channel!

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