LeetCode //C - 216. Combination Sum III

216. Combination Sum III

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:

1 + 2 + 4 = 7

There are no other valid combinations.

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:

1 + 2 + 6 = 9

1 + 3 + 5 = 9

2 + 3 + 4 = 9

There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.

Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

2 <= k <= 9
1 <= n <= 60

From: LeetCode

Link: 216. Combination Sum III

Solution:

Ideas：

Dynamic Capacity: The capacity for storing combinations is initially set and then doubled as needed. This prevents overflow by ensuring there's always enough space for new combinations.
Safe Memory Operations: All memory allocations and accesses are carefully managed to ensure they stay within bounds. This includes checking capacity before adding new combinations and correctly allocating and freeing memory.
Base Case Check: The base case in the backtrack function ensures that we only add valid combinations that meet the criteria (sum equals n and combination length equals k).

Caode:

c 复制代码

void backtrack(int target, int k, int start, int* path, int depth, int*** res, int* resSize, int** colSizes, int* capacity) {
    // Base case: if the combination is complete
    if (depth == k) {
        if (target == 0) {
            // If current path sums up to n, add it to the results
            if (*resSize >= *capacity) {
                // Double the capacity if necessary
                *capacity *= 2;
                *res = realloc(*res, *capacity * sizeof(int*));
                *colSizes = realloc(*colSizes, *capacity * sizeof(int));
            }
            (*res)[*resSize] = (int*)malloc(k * sizeof(int));
            for (int i = 0; i < k; i++) {
                (*res)[*resSize][i] = path[i];
            }
            (*colSizes)[*resSize] = k;
            (*resSize)++;
        }
        return;
    }
    
    for (int i = start; i <= 9; i++) {
        if (i > target) break; // Early termination
        path[depth] = i; // Choose
        backtrack(target - i, k, i + 1, path, depth + 1, res, resSize, colSizes, capacity); // Explore
        // No need to explicitly "unchoose", as path[depth] will be overwritten in the next iteration
    }
}

int** combinationSum3(int k, int n, int* returnSize, int** returnColumnSizes) {
    int capacity = 128; // Initial capacity for results
    int** res = (int**)malloc(capacity * sizeof(int*));
    *returnColumnSizes = (int*)malloc(capacity * sizeof(int));
    *returnSize = 0;

    int* path = (int*)malloc(k * sizeof(int)); // Temp storage for the current combination
    backtrack(n, k, 1, path, 0, &res, returnSize, returnColumnSizes, &capacity);
    
    free(path); // Cleanup
    return res;
}