leetcode 46 不重复数组全排列 47. 全排列 II_int** permute(int* nums, int numssize, int* return-优快云博客

本文链接：https://blog.youkuaiyun.com/chen_ethan/article/details/104225568

本文深入解析了全排列算法的实现细节，包括递归方法和状态记录技巧，通过具体代码示例展示了如何生成一个序列的所有可能排列，适用于无重复数字的序列，并提供了避免重复计算的优化策略。

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题目：给定一个没有重复数字的序列，返回其所有可能的全排列。

易错点：

正确写法： (*returnColumnSizes)[i] = numsSize;

错误写法： *returnColumnSizes[i] = numsSize;

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
void find(int index, int* path, int *history, int *nums, int numsSize, int ** result, int* result_size)
{
    for (int i = 0; i < numsSize; i++) {
	//printf("index:%d %d\n", index, path[index]);
    if (index == numsSize) {
        result[*result_size] = (int *)malloc(numsSize * sizeof(int));
        for (int i = 0; i < numsSize; i++) {
            result[*result_size][i] = path[i];
        }
            *result_size += 1;
            return;
    }
    
        if(history[i] == 1) continue;
		
        path[index++] = nums[i];
        history[i] = 1;
        find(index, path, history, nums, numsSize, result, result_size);
        history[i] = 0;
        index--;
    }
}
int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    if(!nums || numsSize <= 0) {
        *returnSize = 0;
        *returnColumnSizes = NULL;
        return NULL;
    }
    int* history = (int *)calloc(numsSize, sizeof(int));
    int **ret = (int **) calloc(4096, sizeof(int *));
    *returnColumnSizes = (int *)calloc (4096, sizeof(int));
    int path[4096];
    int size = 0;
    find(0, path, history, nums, numsSize, ret, &size);
    for (int i = 0; i < size; i++) {
        (*returnColumnSizes)[i] = numsSize;
    }
    *returnSize = size;
    return ret;
}

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
void find(int index, int* path, int *history, int *nums, int numsSize, int ** result, int* result_size)
{
    printf("index:%d %d\n", index, path[index]);
    if (index == numsSize) {
       result[*result_size] = (int *)malloc(numsSize * sizeof(int));
        for (int i = 0; i < numsSize; i++) {
            result[*result_size][i] = path[i];
            printf("%d ", path[i]);
        }
            printf("\n");
            *result_size += 1;
            return;
    }
    for (int i = 0; i < numsSize; i++) {
        if(history[i] == 1 || (i > 0 && nums[i] == nums[i - 1] && history[i - 1] == 0)) {
            continue;
        }
        path[index++] = nums[i];
        history[i] = 1;
        find(index, path, history, nums, numsSize, result, result_size);
        history[i] = 0;
        index--;
    }
}
int cmp(const void *a, const void *b)
{
    return *((int *)a) - *((int *)b);
}
int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    if(!nums || numsSize <= 0) {
        *returnSize = 0;
        *returnColumnSizes = NULL;
        return NULL;
    }
    qsort(nums, numsSize, sizeof(int), cmp);
    int* history = (int *)calloc(numsSize, sizeof(int));
    int **ret = (int **) calloc(4096, sizeof(int *));
    *returnColumnSizes = (int *)calloc (4096, sizeof(int));
    int path[4096];
    int size = 0;
    find(0, path, history, nums, numsSize, ret, &size);
    for (int i = 0; i < size; i++) {
        (*returnColumnSizes)[i] = numsSize;
    }
    *returnSize = size;
    return ret;
}