【
给定一个可能包含重复元素的整数数组 nums,返回该数组所有可能的子集(幂集)。
说明:解集不能包含重复的子集。
示例:
输入: [1,2,2]
输出:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/subsets-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。】
78题的升级版,多了重复元素如何处理,大概思路同40题,此处不再重复说明。
int rsize = 0;
void dfs(int **retarr, int itend, int* nums, int numsSize, int** returnColumnSizes, int *colnum, int *temp) {
int i;
//if(*colnum > 0) printf("itend=%d,temp[%d]=%d,rsize=%d\n", itend, *colnum-1, temp[*colnum-1],rsize);
returnColumnSizes[0][rsize] = *colnum;
memcpy(retarr[rsize], temp, (*colnum) * sizeof(int));
rsize++;
for (i = itend; i < numsSize; i++) {
temp[*colnum] = nums[i];
*colnum = *colnum + 1;
dfs(retarr, i+1, nums, numsSize, returnColumnSizes, colnum, temp);
*colnum = *colnum - 1;
temp[*colnum] = 0;
while(((i+1) < numsSize) && (nums[i] == nums[i+1])) {
i++;
}
}
}
int comp(const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
int i;
int j;
int t = 1 << numsSize;
int colCount = 0;
int **retarr;
int *temp;
*returnSize = t;
rsize = 0;
returnColumnSizes[0] = (int *)malloc(sizeof(int) * t);
temp = (int *)malloc(sizeof(int) * t);
memset(temp, 0, sizeof(int) * t);
retarr = (int **)malloc(sizeof(int *) * t);
for(i = 0; i < t; i++) {
retarr[i] = (int *)malloc(sizeof(int) * t);
memset(retarr[i], 0, sizeof(int) * t);
}
qsort(nums, numsSize, sizeof(int), comp);
/*for(i = 0; i < numsSize; i++) {
printf("nums[%d]=%d\n", i, nums[i]);
}*/
dfs(retarr, 0, nums, numsSize, returnColumnSizes, &colCount, temp);
*returnSize = rsize;
return retarr;
}