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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 }
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原文地址：http://www.cnblogs.com/bakari/p/4856141.html
