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题目描述:
给定一个可包含重复数字的序列,返回所有不重复的全排列。
示例:
输入: [1,1,2]
输出:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
题目解答:
方法1:回溯
因为有重复元素,所以需要先排序,然后进行回溯。每次将没有添加过的元素加入临时数组,相等的则直接跳过,注意要过滤掉重复元素。
运行时间8ms,代码如下。
/**
* Return an array of arrays of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int comp(const void* a, const void* b) {
return *(int*)a - *(int*)b;
}
void dfs(int*** result, int* size, int* nums, int n, int* before, int bef, bool* flag) {
if(bef == n) {
(*size)++;
result[0] = (int**)realloc(result[0], *size * sizeof(int*));
result[0][*size - 1] = (int*)malloc(n * sizeof(int));
memcpy(result[0][*size - 1], before, n * sizeof(int));
return;
}
int i = 0;
for(i = 0; i < n; i++) {
if(flag[i])
continue;
// if (i > 0 && nums[i] == nums[i - 1] && !flag[i - 1])
// continue; //可以把下边的while循环换成这两句话。
flag[i] = true;
before[bef] = nums[i];
dfs(result, size, nums, n, before, bef + 1, flag);
flag[i] = false;
while(i + 1 < n && nums[i] == nums[i + 1])
i++;
}
}
int** permuteUnique(int* nums, int numsSize, int* returnSize) {
qsort(nums, numsSize, sizeof(int), comp);
int** result = NULL;
int* before = (int*)malloc(numsSize * sizeof(int));
bool* flag = (bool*)calloc(numsSize, sizeof(bool));
dfs(&result, returnSize, nums, numsSize, before, 0, flag);
free(before);
free(flag);
return result;
}
交换法,因为每次交换之后的序列,有可能是无序的,但又有重复元素,所以需要每次交换前都需要先进行排序。
运行时间8ms,代码如下。
/**
* Return an array of arrays of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int comp(const void* a, const void* b) {
return *(int*)a - *(int*)b;
}
void dfs(int*** result, int* size, int* nums, int n, int start) {
if(start + 1 == n) {
(*size)++;
result[0] = (int**)realloc(result[0], *size * sizeof(int*));
result[0][*size - 1] = (int*)malloc(n * sizeof(int));
memcpy(result[0][*size - 1], nums, n * sizeof(int));
return;
}
int i = 0;
int* temp = (int*)malloc(n * sizeof(int));
memcpy(temp, nums, n * sizeof(int));
qsort(nums + start, n - start, sizeof(int), comp);
for(i = start; i < n; i++) {
int t = nums[start];
nums[start] = nums[i];
nums[i] = t;
dfs(result, size, nums, n, start + 1);
nums[i] = nums[start];
nums[start] = t;
while(i + 1 < n && nums[i] == nums[i + 1])
i++;
}
memcpy(nums, temp, n * sizeof(int));
free(temp);
}
int** permuteUnique(int* nums, int numsSize, int* returnSize) {
qsort(nums, numsSize, sizeof(int), comp);
int** result = NULL;
dfs(&result, returnSize, nums, numsSize, 0);
return result;
}