算法导论学习笔记(3)－习题2.3-7－排序＋二分

question(题意)：

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

设计一个O(n lg(n))时间复杂度的算法，给定一个n个数字的集合，和一个数字x, 问，是否能在这个集合中找到两个数字，他们的和恰好为x.

Solution(解法)：

在一个数组里面找某个数，O(n lg(n))，很容易想到用排序+二分，算法导论刚刚讲了归并排序，正好是O(n lg(n))的排序算法，可是这个题是找两个数的和为x, 那么该如何做呢，我们可以在排序后的数组里面，从小到大一次枚举第一个数（该数字必须满足<=x/2,原因自己想)，然后在后面查找是否有另外一个数即可，下面是代码：

Code(代码):

//算法导论 习题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, 10, x))
        {
            cout << "find two numbers whose sum is " << x << endl;
        }
        else
        {
            cout << "Not find two numbers whose sum is "<< x << endl;
        }
    }
    
    return 0;
}