本篇博文意在对前几章中遗漏的，本人觉得有意思的习题当独拿出来练练手。
1、习题2-4，求逆序对，时间复杂度要求Θ(nlgn)
定义：对于一个有n个不同的数组A, 当i<j时，存在A[i]>A[j]，则称对偶(i, j)为A的一个逆序对。
譬如：<2,3,8,6,1>有5个逆序对。
解题思路：归并排序的思想：逆序对的数量=左区间的逆序对+右区间的逆序对+合并的逆序对
代码如下：
1 #include <iostream>
2 #include <vector>
3 using namespace std;
4
5 int MergeInversion(int arr[], int nLeft, int nRight);
6 int Merge(int arr[], int p, int q, int r);
7
8 //用归并排序找逆序对
9 int MergeInversion(int arr[], int nLeft, int nRight)
10 {
11 int nInversion = 0;
12 if (nLeft < nRight) {
13 int nMid = (nLeft+nRight) >> 1;
14 nInversion += MergeInversion(arr, nLeft, nMid);
15 nInversion += MergeInversion(arr, nMid+1, nRight);
16 nInversion += Merge(arr, nLeft, nMid, nRight);
17 }
18 return nInversion;
19 }
20
21 int Merge(int arr[], int p, int q, int r)
22 {
23 int n1 = q - p + 1;
24 int n2 = r - q;
25 int *left = new int[n1];
26 int *right = new int[n2];
27
28 int i,j;
29 for (i = 0; i < n1; i ++)
30 left[i] = arr[p+i];
31 for (j = 0; j < n2; j ++)
32 right[j] = arr[q+1+j];
33
34 i = 0;
35 j = 0;
36 int k = p;
37 int nInverCount = 0; //逆序对的数目
38
39 while (i < n1 && j < n2) {
40 if (left[i] <= right[j])
41 arr[k++] = left[i++];
42 else {//假如左边子序列的数大于右子序列，则逆序对数为n1 - i;
43 arr[k++] = right[j++];
44 nInverCount += (n1 - i);
45 }
46 }
47 while (i < n1)
48 arr[k++] = left[i++];
49 while (j < n2)
50 arr[k++] = right[j++];
51
52 delete left;
53 left = NULL;
54 delete right;
55 right = NULL;
56
57 return nInverCount;
58 }
59
60 int main()
61 {
62 int arr[] = {2,3,8,6,1};
63 int len = sizeof(arr)/sizeof(arr[0]);
64
65 int *arrT = new int[len];
66
67
68 cout << MergeInversion1(arr, 0, 4, arrT) << endl;
69 for (int i = 0; i < 5; i++)
70 cout << arrT[i] << " ";
71 return 0;
72 }
2、习题9.3-7，设计一个O(n)时间的算法，对于一个给定的包含n个互异元素的集合S和一个正整数 k<=n，该算法能够确定S中最接近中位数的k个元素。
譬如数组：<5,7,10,3,6,2,8,9,1,11,12>，中位数为6，如果k=3，最接近6的3个元素是<5,7,4>或<5,7,8>；如果k=4，则最接近6的4个元素是<5,7,4,8>。
解题思路：
1）通过Select()得到A的中位数 --->O(n)
2）计算A中每个数到中位数的差值作为数组D ，并增加D的一个副本D‘ --->O(n)
3）通过Select()得到D’中第k大的数 ---> O(n)
4）依次遍历D中的数，判断比k小的k个数，即为所求 ---> <O(n)
代码如下：
1 #include <iostream>
2 #include <cassert>
3 #include <cmath>
4 using namespace std;
5
6 int* K_Select(int arr[], int nLen, int k);
7 int Select(int arr[], int nLeft, int nRight, int nMin);
8 int Partition(int arr[], int p, int r);
9 void Swap(int &n, int &m);
10
11 int* K_Select(int arr[], int nLen, int k)
12 {
13 assert(k <= nLen);
14
15 int *pDis = new int[nLen-1]; //the distance of every bit and median
16 int *pDis_cpy = new int[nLen-1];
17 int *pKArr = new int[k]; //return k_array
18
19 int nMedian = Select(arr, 0, nLen-1, nLen/2); //get the median of arr
20 int iMedian;
21 int nDLen = 0;
22 for (int i = 0; i < nLen; i ++) {
23 if (arr[i] != nMedian)
24 pDis[nDLen++] = abs(arr[i]-nMedian); //get the distance using abs()
25 else
26 iMedian = i; //get the index of median
27 }
28 memcpy(pDis_cpy, pDis, sizeof(int)*(nLen-1)); //copy the distance
29 int nKMedian = Select(pDis_cpy, 0, nDLen-1, k); //get the k-th minimum distance between median and every bit
30
31 delete pDis_cpy;
32 pDis_cpy = NULL;
33
34 int ik = 0;
35 for (int i = 0; ik < k && i < nDLen; i ++) {
36 if (pDis[i] <= nKMedian) {
37 if (i < iMedian) //judge the index of array and the index of median
38 pKArr[ik++] = nMedian - pDis[i];
39 else
40 pKArr[ik++] = nMedian + pDis[i];
41 }
42 }
43
44 delete pDis;
45 pDis = NULL;
46
47 return pKArr;
48 }
49
50 int Select(int arr[], int nLeft, int nRight, int nMin)
51 {
52 assert(nLeft <= nRight);
53 assert(nMin <= nRight-nLeft+1);
54
55 if (nLeft == nRight)
56 return arr[nLeft];
57 int nMid = Partition(arr, nLeft, nRight);
58 int k = nMid - nLeft + 1;
59 if (k == nMin)
60 return arr[nMid];
61 else if (k > nMin)
62 return Select(arr, nLeft, nMid-1,nMin);
63 else
64 return Select(arr, nMid+1, nRight, nMin-k);
65 }
66
67 int Partition(int arr[], int p, int r)
68 {
69 assert(p <= r);
70
71 int nTemp = arr[r];
72 int i = p - 1, j = p;
73 while(j < r) {
74 if (arr[j] <= nTemp) {
75 Swap(arr[i+1], arr[j]);
76 i ++;
77 }
78 j ++;
79 }
80 Swap(arr[i+1], arr[r]);
81 return i+1;
82 }
83
84 void Swap(int &n, int &m)
85 {
86 int t = n;
87 n = m;
88 m = t;
89 }
90
91 int main()
92 {
93 int arr[] = {5,7,10,3,6,2,8,9,4,1,11,12};
94 int *karr = K_Select(arr, 12, 4);
95 for (int i=0; i < 4; i ++)
96 cout << karr[i] << " ";
97 return 0;
98 }
3、习题9.3-8，求两个有序数组的中位数？
题目：设X[1...n]和Y[1...n]为两个数组，每个都包含n个有序的元素。请设计一个O(lgn)时间的算法来找出数组X 和Y中所有2n个元素的中位数。
譬如：X：<1,2,3,4,7,9>; Y:<2,5,6,7,10,11>; 中位数为数组Y中的6.
解题思路：二分查找思想--->O(lgn)
1）如果两个数组均只有一个元素，则返回数组中的较小值。
2）两个数组有2n个元素，则中位数位于n index，小于中位数的元素个数为n-1;
3）如果当前找到中位数在数组X中的标号为 K，则在数组Y的标号为：n-K-2 < index <= n-K-1;
4）如果当前在数组X中找到的中位数 X[K]<= Y[n-K-2]，说明该数小了，应该在X的右区间[mid+1, right]继续找，反之，如果X[K] > Y[n-K-1]，说明该数大了，应该在X的左区间[left, mid-1]继续找。
5）如果重复该过程，没有在X中找到该中位数，则采用同样的方法在Y中找。
代码如下：
1 #include <iostream>
2 #include <cassert>
3 using namespace std;
4
5 int SelectMedian(int *arrA, int *arrB, int left, int right)
6 {
7 assert(NULL != arrA);
8 assert(NULL != arrB);
9
10 int nLen = right-left+1;
11
12 while (left <= right) {
13 int mid = (left+right)/2;
14
15 if (mid == nLen-1 && arrA[mid] <= arrB[0]) //only one element
16 return arrA[mid];
17 else if (mid < nLen-1) {
18 if(arrA[mid] <= arrB[nLen-mid] && arrA[mid] > arrB[nLen-1-mid]) //K = nLen-1-K note:from 0 index
19 return arrA[mid];
20 else if (arrA[mid] <= arrB[nLen-1-mid]) //if the element is small, find it in right region.
21 left = mid + 1;
22 else //if the element is big, find it in left region.
23 right = mid - 1;
24 }
25 }
26 return -1;
27 }
28
29 //find the median of two array: arrA and arrB
30 //return the median of two array
31 void TwoArrMedian(int *arrA, int *arrB, int nLen)
32 {
33 assert(NULL != arrA);
34 assert(NULL != arrB);
35
36 int nMedian;
37 if (SelectMedian(arrA, arrB, 0, nLen-1) == -1) //not find
38 nMedian = SelectMedian(arrB, arrA, 0, nLen-1);
39
40 cout << nMedian << endl;
41 }
42
43 int main()
44 {
45 int a[] = {1,2,3,4,8,9,13};
46 int b[] = {2,5,6,7,10,11,12};
47
48 TwoArrMedian(a, b, 7);
49 return 0;
50 }
时间： 2024-10-05 07:00:53