如何实现在n个数里找到两个数和为sum

METHOD 1 (Use Hashing)

// C++ program to check if given array 
// has 2 elements whose sum is equal 
// to the given value 

// Works only if range elements is limited 
#include <stdio.h> 
#define MAX 100000 

void printPairs(int arr[], int arr_size, int sum) 
{ 
int i, temp; 
bool s[MAX] = {0}; /*initialize hash set as 0*/

for (i = 0; i < arr_size; i++) 
{ 
    temp = sum - arr[i]; 
    if (temp >= 0 && s[temp] == 1) 
        printf("Pair with given sum %d is (%d, %d) n", 
                sum, arr[i], temp); 
    s[arr[i]] = 1; 
} 
} 

/* Driver program to test above function */
int main() 
{ 
    int A[] = {1, 4, 45, 6, 10, 8}; 
    int n = 16; 
    int arr_size = sizeof(A)/sizeof(A[0]); 

    printPairs(A, arr_size, n); 

    getchar(); 
    return 0; 
} 

METHOD 1 (Use Sorting)

// C program to check if given array 
// has 2 elements whose sum is equal 
// to the given value 

# include <stdio.h> 
# define bool int 

void quickSort(int *, int, int); 

bool hasArrayTwoCandidates(int A[], int arr_size, int sum) 
{ 
    int l, r; 

    /* Sort the elements */
    quickSort(A, 0, arr_size-1); 

    /* Now look for the two candidates in the sorted 
    array*/
    l = 0; 
    r = arr_size-1; 
    while (l < r) 
    { 
        if(A[l] + A[r] == sum) 
            return 1; 
        else if(A[l] + A[r] < sum) 
            l++; 
        else // A[i] + A[j] > sum 
            r--; 
    }     
    return 0; 
} 

/* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING 
    PURPOSE */
void exchange(int *a, int *b) 
{ 
    int temp; 
    temp = *a; 
    *a = *b; 
    *b = temp; 
} 

int partition(int A[], int si, int ei) 
{ 
    int x = A[ei]; 
    int i = (si - 1); 
    int j; 

    for (j = si; j <= ei - 1; j++) 
    { 
        if(A[j] <= x) 
        { 
            i++; 
            exchange(&A[i], &A[j]); 
        } 
    } 
    exchange (&A[i + 1], &A[ei]); 
    return (i + 1); 
} 

/* Implementation of Quick Sort 
A[] --> Array to be sorted 
si --> Starting index 
ei --> Ending index 
*/
void quickSort(int A[], int si, int ei) 
{ 
    int pi; /* Partitioning index */
    if(si < ei) 
    { 
        pi = partition(A, si, ei); 
        quickSort(A, si, pi - 1); 
        quickSort(A, pi + 1, ei); 
    } 
} 

/* Driver program to test above function */
int main() 
{ 
    int A[] = {1, 4, 45, 6, 10, -8}; 
    int n = 16; 
    int arr_size = 6; 
    
    if( hasArrayTwoCandidates(A, arr_size, n)) 
        printf("Array has two elements with given sum"); 
    else
        printf("Array doesn't have two elements with given sum"); 

    getchar(); 
    return 0; 
}