一、491 非递减子序列

本题不能按照前面的方法来判断重复数字，因为数组本来就带顺序属性，不能移动。

主要在于怎么选择子序列：首先同层的处理，需要判断，当前层前面出现过的数字，后面不能再使用（很重要！！）；另外，如果path头部数字和当前数字不是递增的，那么需要跳过。

int* path;int** result; //int used[201];int path_index;int resize;int* recolsize;void backtracking(int* nums, int numsSize, int startidx)
{if(path_index > 1){result[resize] = (int*)malloc(sizeof(int) * path_index);memcpy(result[resize], path, path_index * sizeof(int));recolsize[resize] = path_index;resize++;}int used[201] = {0};for(int i = startidx; i < numsSize; i++){/**不满足条件的情况**/if((path_index > 0 && nums[i] < path[path_index -1]) || (used[nums[i] + 100] == 1))//path[path_index -1]是top元素{continue;}path[path_index++] = nums[i];used[nums[i] + 100] = 1;backtracking(nums, numsSize, i+ 1);path_index--;}
}int** findSubsequences(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {path = (int*)malloc(sizeof(int) * 20);result = (int**)malloc(sizeof(int*) * 50000);recolsize = (int*)malloc(sizeof(int) * 50000);path_index = 0;resize = 0;*returnSize = 0;*returnColumnSizes = NULL;if(numsSize <= 0){return NULL;}backtracking(nums, numsSize, 0);*returnSize = resize;*returnColumnSizes = recolsize;return result;
}

二、46 全排列

全排列是找所有可能的组合，因此，不需要用startindx来拆分了，数字都会重复使用，但是在纵向迭代时候，需要记录这个值是否使用过，如果使用过，那么跳出；

int** result;
int* path;
int path_idx;
int* used;
int resize;
int* recolsize;void backtracking(int* nums, int numsSize, int* used)
{if(path_idx == numsSize){result[resize] = (int*)malloc(sizeof(int) * path_idx);recolsize[resize] = path_idx;memcpy(result[resize], path, path_idx * sizeof(int));resize++;return;}for(int i = 0; i < numsSize; i++){if(used[i] == 1){continue;}path[path_idx++] = nums[i];used[i] = 1;backtracking(nums, numsSize, used);used[i] = 0;path_idx--;}}int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {result = (int*)malloc(sizeof(int*) * 50000);path = (int*)malloc(sizeof(int) * 10);int* used = (int*)calloc(10, sizeof(int));//malloc(sizeof(int) * 10);path_idx = 0;resize = 0;recolsize = (int*)malloc(sizeof(int) * 50000);*returnSize = 0;*returnColumnSizes = NULL;if(numsSize <= 0){return NULL;}backtracking(nums, numsSize, used);*returnColumnSizes = recolsize;*returnSize = resize;return result;}

三、全排列II

 主要是理解used在树枝和树层的含义！

void sortArr(int* arr, int size)
{for(int i = 0; i < size; i++){for(int j = i + 1; j < size; j++){if(arr[i] > arr[j]){int tmp = 0;tmp = arr[i];arr[i] = arr[j];arr[j] = tmp;}}}
}
int** result;
int* path;
int path_idx;
int* used;
int resize;
int* recolsize;// used[i - 1] == true，说明同一树枝nums[i - 1]使用过// used[i - 1] == false，说明同一树层nums[i - 1]使用过// 如果同一树层nums[i - 1]使用过则直接跳过
void backtracking(int* nums, int numsSize, int* used)
{if(path_idx == numsSize){result[resize] = (int*)malloc(sizeof(int) * path_idx);recolsize[resize] = path_idx;memcpy(result[resize], path, path_idx * sizeof(int));resize++;return;}for(int i = 0; i < numsSize; i++){if(i > 0 && (used[i -1] == 0) && (nums[i] == nums[i -1])){continue;}if(used[i] == 1){continue;}path[path_idx++] = nums[i];used[i] = 1;backtracking(nums, numsSize, used);used[i] = 0;path_idx--;}}int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {result = (int*)malloc(sizeof(int*) * 50000);path = (int*)malloc(sizeof(int) * 10);int* used = (int*)calloc(10, sizeof(int));//malloc(sizeof(int) * 10);path_idx = 0;resize = 0;recolsize = (int*)malloc(sizeof(int) * 50000);*returnSize = 0;*returnColumnSizes = NULL;if(numsSize <= 0){return NULL;}sortArr(nums, numsSize);backtracking(nums, numsSize, used);*returnColumnSizes = recolsize;*returnSize = resize;return result;
}