题解 | #有重复项数字的全排列#

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param num int整型一维数组 
 * @param numLen int num数组长度
 * @return int整型二维数组
 * @return int* returnSize 返回数组行数
 * @return int** returnColumnSizes 返回数组列数
 */
 void back(
        int* num,
        int numLen,
        int** res,
        int* path,
        int pathsize,
        int* used,
        int* len       
    ){
        if(pathsize == numLen){
            int *temp = (int *)malloc(sizeof(int) * numLen);
            memcpy(&temp[0], &path[0], sizeof(int) * numLen);
            res[len[0]] = temp;
            len[0]++;
            return;
        }
        for(int i = 0 ; i < numLen ; i++){
            if(i != 0 && num[i - 1] == num[i] && used[i - 1] == 0) continue;
            if(used[i] == 1) continue;
            path[pathsize] = num[i];
            used[i] = 1;
            back(num,numLen,res,path,pathsize + 1,used,len);
            used[i] = 0;
        }
    }
    int cmp(const void* a , const void* b){
        return *(int*)a - *(int*)b;
    }
int** permuteUnique(int* num, int numLen, int* returnSize, int** returnColumnSizes ) {
    // write code here
   if(numLen == 0) return NULL;
    int* path = (int*)malloc(sizeof(int)*numLen);
    int* used = (int*)malloc(sizeof(int)*numLen);
    int** res = (int**)malloc(sizeof(int*)*1000);
    int len = 0;
    for(int i = 0 ; i < numLen ; i++){
        used[i] = 0;
    }
    qsort(num , numLen , sizeof(int) , cmp);
     back(num,numLen,res,path,0,used,&len);
     printf("%d\n" , len);
     *returnSize = len;
    *returnColumnSizes = (int*)malloc((*returnSize) * sizeof(int));
    for(int i = 0; i < *returnSize; i++) {
        (*returnColumnSizes)[i] = numLen;
    }
     free(path);
     free(used);
    return res;
}