Introduction to Algorithms Learning Notes (3)-Exercise 2.3-7-sorting + two points

Question (test instructions):

Describe a O (n LG (n))-time algorithm, given a set S of n integers and another integer x, determines whether or not th ere exist elements in S whose sum is exactly x.

Design an O (n LG (n)) time complexity algorithm, given a set of n numbers, and a number x, asking if it is possible to find two numbers in this set, their and exactly X.

Solution (Solution):

In an array to find a certain number, O (n LG (n)), it is easy to think of sorting + two points, the introduction of the algorithm has just talked about the merge sort, is just the O (n LG (n)) sorting algorithm, but this question is to find two number of and X, then how to do it, we can in the sorted array From small to large enumerate the first number (the number must satisfy <=X/2, reason to think), and then find out if there is another number, the following is the code:

Code (CODES):

Introduction to Algorithmic exercises 2.3-7#include <iostream>using namespace std;void merge (int* A, int p, int q, int r) {int N1 = Q-p +1,    N2 = R-q;        int l[n1], r[n2];    for (int i = 0; i < N1; i++) l[i] = A[p+i];    for (int i = 0; i < n2; i++) r[i] = a[q+i+1];    int i = 0, j = 0, k = p;            while (K <= R) {if (I >= N1) {while (J < N2) {a[k++] = r[j++];            }} else if (J >= N2) {while (I < N1) {a[k++] = l[i++];        }} else if (L[i] <= r[j]) {a[k++] = l[i++];        } else {a[k++] = r[j++];        }}}void Merge_sort (int* A, int p, int r) {if (P < r) {int q = (p + r)/2;        Merge_sort (A, p, q);        Merge_sort (A, q+1, R);    Merge (A, p, Q, R);        }}bool bsearch (int* A, int L, int R, int v) {while (L <= r) {int M = (L + r)/2;        if (A[m] < v) {L = m+1;        }else if (A[m] > V) {R = M-1;        } else {return true; }} return false;}        BOOL Find (int* A, int n, int x) {merge_sort (A, 0, n-1);                for (int i = 0; i < n; i++) {if (A[i] <= x/2) {if (bsearch (A, i+1, N-1, X-a[i])) {            return true;        }} else {return false; }} return false;}        int main () {int a[10] = {2, 3, 1, 5, 3, 2, 9, 3, 0, 7}, X; while (CIN >> x) {if (Find (A, ten, X)) {cout << "find" numbers whose sum is "        << x << Endl;        } else {cout << ' not ' find numbers whose sum is ' << x << Endl; }} return 0;}

Introduction to Algorithms Learning Notes (3)-Exercise 2.3-7-sorting + two points