Leetcode 题解 Combinations：回溯+求排列组合

罗列出从n中取k个数的组合数组。

首先，求C(n,k)这个实现，很粗糙，溢出也不考虑，好的方法也不考虑。笨蛋。心乱，上来就写。。

另外，发现在递归中，不能申请太大的数组？貌似不是这个问题，是我自己越界了。

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int com(int n,int k)
{
    int i = ,ans=;

    if(k > n-k) k = n-k;

    for (i = n; i > n - k; i--)
        ans *= i;
    for (i = k; i > ; i--)
        ans /= i;
    return ans;
}

int **ans;
int len;

void back_track(int *last, int last_size, int num, int n, int k)
{
    if(last_size == k)
    {
        int *tmp = (int*)malloc(sizeof(int)*k);
        memcpy(tmp,last,k*sizeof(int));
        ans[len++] = tmp;
        //memcpy(ans[len],last,k*sizeof(int));
        //len ++;
        return;
    }

    if(num > n) return;

    back_track(last,last_size,num+,n,k);
    last[last_size] = num;
    back_track(last,last_size + ,num + ,n,k);
}
int** combine(int n, int k, int** columnSizes, int* returnSize) {
    if(k > n || n *k == )  { *returnSize = ;  return NULL; }

    int size = com(n,k);
    int *cols = (int*)malloc(size*sizeof(int));

    ans = (int**)malloc(size*sizeof(int*));
    len = ;

    for(int i = ; i < size; i ++)
    {
        //ans[i] = (int*)malloc(k*sizeof(int));
        cols[i] = k;
    }

    int *last = (int*)malloc(k*sizeof(int));
    back_track(last,,,n,k);

    *columnSizes = cols;
    *returnSize = len;
    return ans;
}