这个问题更简单或更流行的版本是找到具有给定总和的三元组。但是这个提出了一个额外的条件。找到未排序数组中的所有三元组，使得

d[i]+d[j]+d[k] <= t;   & d[i]<d[j]<d[k] where i<j<k

最佳答案

求求和小于或等于 k ​​的递增三元组:

# include <stdio.h>
void find3Numbers(int A[], int arr_size, int sum)
{
    int l, r;
    for (int i = 0; i < arr_size-2; i++){
       for (int j = i+1; j < arr_size-1; j++){         
           for (int k = j+1; k < arr_size; k++){
               if (A[i] + A[j] + A[k] <= sum)
                 printf("Triplet is %d, %d, %d\n", A[i], A[j], A[k]);
            }
        }
     }
}
int main()
{
    int A[] = {1, 2, 3, 4, 6};
    int sum = 8;
    int arr_size = sizeof(A)/sizeof(A[0]);
    find3Numbers(A, arr_size, sum);
    return 0;
}

Execution :
arr_size = 5
Step:1   i=0 and i<3 (arr_size-2)
                                j=1 and j<4 (arr_size-1)
                                                k=2 and k<5 (arr_size)
                                                                A[0]+A[1]+A[2]<=sum --> 1+2+3 <=8 --> 6<=8 ( true )
                                                k=3 and k<5
                                                                A[0]+A[1]+A[3]<=sum --> 1+2+4 <=8 --> 7<=8 ( true )
                                                k=4 and k<5
                                                                A[0]+A[1]+A[4]<=sum --> 1+2+6 <=8 --> 9<=8 ( false )
                                j=2 and j<4
                                                k=3 and k<5
                                                                A[0]+A[2]+A[3]<=sum --> 1+3+4 <=8 --> 8<=8 ( true )
                                                k=4 and k<5
                                                                A[0]+A[2]+A[4]<=sum --> 1+3+6 <=8 --> 10<=8 ( false )
                                j=3 and j<4
                                                k=4 and k<5
                                                                A[0]+A[3]+A[4]<=sum --> 1+4+6 <=8 --> 11<=8 ( false )
                                j=4 and j<4 (false)
Step:2  i=1 and i<3
                                j=2 and j<4
                                                k=3 and k<5
                                                                A[1]+A[2]+A[3]<=sum --> 2+3+4 <=8 --> 9<=8 ( false )
                                                k=4 and k<5
                                                                A[1]+A[2]+A[4]<=sum --> 2+3+6 <=8 --> 11<=8 ( false )
                                j=3 and j<4
                                                k=4 and k<5
                                                                A[1]+A[3]+A[4]<=sum --> 2+4+6 <=8 --> 12<=8 ( false )
                                j=4 and j<4 (false)
Step:3 i=2 and i<3
                                j=3 and j<4
                                                k=4 and k<5
                                                                A[2]+A[3]+A[4]<=sum --> 3+4+6 <=8 --> 13<=8 ( false )
                                j=4 and j<4 (false)
Step:4 i=3 and i<3 (false)

关于java - 找到递增的三元组，使得总和小于或等于 k，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28050746/