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BackTracking_Fixed sum for array elements

 

Given an array a contains distinct positive integers, count how many combinations of integers in a add up to exactly sum

For example, given int[] a = {11, 3, 8} ; and sum = 11

You should output 2, because 11 == 11 and 3 + 8 == 11

This is typically a backtracking problem

Enumerate all the subsets of the given array to see how many of them match the condition

when you write backtracking procedure using recursion, please be careful of which condition do you

use to terminate the loop, in this code snippet, there two conditions,

1. sum == 0

2. t == a.Length

and when t == a.Length, we might be got an solution yet, don't forget this case.

Backtracking

Code

recursive way

Code

C法

复制代码
#include<stdio.h>
#include<stdlib.h>

int count = 0; // number of solutions

/*
 * array - positive numbers
 * n     - element count in array
 * sum   - pair of sum
 * t     - recursion deep
 */
void find_combinations(int *array, int n, int sum, int t) {
    if (t == n) {
        if (sum == 0) {
            count++;
        }
        return;
    }

    if (sum == 0) { // Find a solution
        count++;
    }
    else {
        if (sum >= array[t]) {  // left tree
            find_combinations(array, n, sum - array[t], t + 1);
        }
        if (sum > 0) {                  // right tree
            find_combinations(array, n, sum, t + 1);
        }
    }
}

int main(void) {
    int a[] = {11, 3, 8, 4, 1, 7};
    find_combinations(a, 6, 11, 0);
    printf("%d\n", count);

    system("pause");
    return 0;
}
复制代码

==

 

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