8.3交換排序
2.編寫雙向冒泡算法，在正反兩個方向交替進行掃描，即第一趟把關鍵字最大的元素放在序列的最后面，第二趟把關鍵字最小的元素放在序列的最前面，如此反復進行。

#include <iostream> #include <algorithm> using namespace std;typedef int ElemType;ElemType a[50];void BubbleSort(ElemType a[], int n) {int low = 0, high = n - 1, i;bool flag = true;while (low < high && flag) {flag = false;for (i = low; i < high; i++) {if (a[i] > a[i + 1]) {swap(a[i], a[i + 1]);flag = true;}}high--;for (i = high; i > low; i--) {if (a[i] < a[i - 1]) {swap(a[i], a[i - 1]);flag = true;}}low++;} }int main() {int n;cin >> n;for (int i = 0; i < n; i++) cin >> a[i];BubbleSort(a, n);for (int i = 0; i < n; i++) cout << a[i] << " ";cout << endl;return 0; }

3.已知線性表按順序存儲，且每個元素都是不相同的整形元素，設計把所有奇數移動到所有偶數前邊的算法（要求時間最少，輔助空間最少）。

#include <iostream> #include <algorithm> using namespace std;typedef int ElemType;ElemType A[50];void move(ElemType A[], int len) {int i = 0, j = len - 1;while (i < j) {while (i < j && A[i] % 2 != 0) i++;while (i < j && A[j] % 2 != 1) j--;if (i < j) {swap(A[i], A[j]);i++; j--;}} }int main() {int n;cin >> n;for (int i = 0; i < n; i++) cin >> A[i];move(A, n);for (int i = 0; i < n; i++) cout << A[i] << " ";cout << endl;return 0; }

4.試重新編寫考點精析中的快速排序的劃分算法，使之美詞選取的樞軸值都是隨機地從當前子表中選擇的。

#include <cstdlib> #include <iostream> using namespace std;typedef int ElemType;ElemType A[50];int Partition2(ElemType A[], int low, int high) {int rand_Index = low + rand() % (high - low);swap(A[rand_Index], A[low]);ElemType pivot = A[low];int i = low;for (int j = low + 1; j <= high; j++)if (A[j] < pivot) swap(A[++i], A[j]);swap(A[i], A[low]);return i; }void QuickSort(ElemType A[], int low, int high) {if (low < high) {int pivotpos = Partition2(A, low, high);QuickSort(A, low, pivotpos - 1);QuickSort(A, pivotpos + 1, high);} }int main() {int n;cin >> n;for (int i = 1; i <= n; i++) cin >> A[i];QuickSort(A, 1, n);for (int i = 1; i <= n; i++) cout << A[i] << " ";cout << endl;return 0; }

5.試編寫一個算法，使之能夠在數組L[1…n]中找出第k小的元素（即從小到大排序后處于第k個位置的元素）。

#include <cstdlib> #include <iostream> #include <algorithm> using namespace std;int a[50];int kth_elem(int a[], int low, int high, int k) {int rand_Index = low + rand() % (high - low + 1);swap(a[rand_Index], a[low]);int pivot = a[low];int low_temp = low; int high_temp = high;while (low < high) {while (low < high && a[high] >= pivot) --high;a[low] = a[high];while (low < high && a[low] <= pivot) ++low;a[high] = a[low];}a[low] = pivot;if (low == k) return a[low];else if (low > k) return kth_elem(a, low_temp, low - 1, k);else return kth_elem(a, low + 1, high_temp, k); }int main() {int n, k;cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];cin >> k;int ans = kth_elem(a, 1, n, k);cout << ans << endl;return 0; }

6.已知由$n(n\ge 2)$個正整數構成的集合$A=a_k|0\le k<n$，將其劃分為兩個不相交的子集$A_1$和$A_2$，元素個數分別是$n_1$和$n_2$，$A_1$和$A_2$中的元素之和分別為$S_1$和$S_2$。設計一個盡可能高效的劃分算法，滿足$|n_1?n_2|$最小且$|S_1?S_2|$最大。

#include <iostream> using namespace std;int a[50];int setPartition(int a[], int n) {int pivotkey, low = 0, low0 = 0, high = n - 1, high0 = n - 1, flag = 1, k = n / 2, i;int s1 = 0, s2 = 0;while (flag) {pivotkey = a[low];while (low < high) {while (low < high && a[high] >= pivotkey) --high;if (low != high) a[low] = a[high];while (low < high && a[low] <= pivotkey) ++low;if (low != high) a[high] = a[low];}a[low] = pivotkey;if (low == k - 1) flag = 0;else {if (low < k - 1) {low0 = ++low;high = high0;} else {high0 = --high;low = low0;}}}for (i = 0; i < k; i++) s1 += a[i];for (i = k; i < n; i++) s2 += a[i];return s2 - s1; }int main() {int n;cin >> n;for (int i = 0; i < n; i++) cin >> a[i];int ans = setPartition(a, n);cout << ans << endl;return 0; }

7.荷蘭國旗問題：設有一個僅有紅、白、藍三種顏色的條塊組成的條塊序列，請編寫一個時間復雜度為$Q(n)$的算法，使得這些條塊按紅、白、藍的順序排好，即排成荷蘭國旗圖案。

#include <cstdlib> #include <iostream> #include <algorithm> using namespace std;char a[50];void Flag_Arrange(char a[], int n) {int i = 0, j = 0, k = n - 1;while (j <= k) {switch (a[j]) {case 'R': swap(a[i], a[j]); i++; j++; break;case 'W': j++; break;case 'B': swap(a[j], a[k]); k--;}} }int main() {cin >> a;int n = strlen(a);Flag_Arrange(a, n);for (int i = 0; i < n; i++) cout << a[i];return 0; }

8.4選擇排序
4.編寫一個算法，在基于單鏈表表示的帶排序關鍵字序列上進行簡單選擇排序。

#include <stdio.h> #include <stdlib.h>typedef int ElemType;typedef struct LinkNode {ElemType data;struct LinkNode *link; } LinkNode, *LinkedList;LinkedList init() {LinkedList head = (LinkedList)malloc(sizeof(LinkNode));int n;ElemType x;scanf("%d", &n);scanf("%d", &x);head->data = x;head->link = NULL;LinkNode *tail = head;for (int i = 1; i < n; i++) {LinkedList temp = (LinkedList)malloc(sizeof(LinkNode));scanf("%d", &x);temp->data = x;tail->link = temp;temp->link = NULL;tail = tail->link;}return head; }void selectSort(LinkedList &L) {LinkNode *h = L, *p, *q, *r, *s;L = NULL;while (h != NULL) {p = s = h; q = r = NULL;while (p != NULL) {if (p->data > s->data) {s = p;r = q;}q = p;p = p->link;}if (s == h) h = h->link;else r->link = s->link;s->link = L;L = s;} }void printList(LinkedList L) {while (L) {printf("%d ", L->data);L = L->link;}printf("\n"); }int main() {LinkedList L = init();selectSort(L);printList(L);return 0; }

5.試設計一個算法，判斷一個數據序列是夠構成一個小根堆。

#include <iostream> using namespace std;typedef int ElemType;ElemType A[50];bool IsMinHeap(ElemType A[], int len) {if (len % 2 == 0) {if (A[len / 2] > A[len]) return false;for (int i = len / 2 - 1; i >= 1; i--)if (A[i] > A[2 * i] || A[i] > A[2 * i + 1]) return false;} else {for (int i = len / 2; i >= 1; i--)if (A[i] > A[2 * i] || A[i] > A[2 * i + 1]) return false;}return true; }int main() {int n;cin >> n;for (int i = 1; i <= n; i++) cin >> A[i];if (IsMinHeap(A, n)) cout << "Yes" << endl;else cout << "No" << endl;return 0; }

8.6各種內部排序算法的比較與應用
4.設有一個數組中存放了一個無序的關鍵序列$K_1,K_2,\dots ,K_n$。現要求將$K_n$放在將元素排序后的正確位置上，試編寫實現該功能的算法，要求比較關鍵字的次數不超過$n$。

#include <iostream> using namespace std;typedef int ElemType;ElemType K[20];int Partition(ElemType K[], int n) {int i = 1, j = n;ElemType pivot = K[j];while (i < j) {while (i < j && K[i] <= pivot) i++;if (i < j) K[j] = K[i];while (i < j && K[j] >= pivot) j--;if (i < j) K[i] = K[j];}K[i] = pivot;return i; }int main() {int n;cin >> n;for (int i = 1; i <= n; i++) cin >> K[i];int p = Partition(K, n);cout << p << endl;return 0; }

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