四种语言刷算法之47. 全排列 II

47. 全排列 II

1、C

/**
 * 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 back(int* nums, int numsSize, int* returnSize, int** returnColumnSizes,int *path,int *pathSize,int **result,int *visited){
    if(numsSize==*pathSize){
        result[*returnSize] = (int *)malloc(sizeof(int)*numsSize);
        memcpy(result[*returnSize],path,sizeof(int)*numsSize);
        (*returnColumnSizes)[*returnSize] = numsSize;
        (*returnSize)++;
        return;
    }
    for(int i=0;i<numsSize;i++){
        if(visited[i]==1||(i>0&&nums[i-1]==nums[i]&&visited[i-1]==0)){
            continue;
        }
        path[*pathSize] = nums[i];
        visited[i] = 1;
        (*pathSize)++;
        back(nums,numsSize,returnSize,returnColumnSizes,path,pathSize,result,visited);
        visited[i] = 0;
        (*pathSize)--;
    }
}
void QuickSort1(int* a, int left, int right)
{
    if (left >= right)
    {
        return;
    }

    int begin = left, end = right;
    //三数取中
    //int midIndex = GetThreeMid(a,begin,end);
    //Swap(&a[begin],&a[midIndex]);
    int pivot = begin;
    int key = a[begin];

    while (begin < end)
    {
        //右边找小的，如果不是小于key,继续
        while (begin < end && a[end] >= key)
        {
            end--;
        }
        //找到比key小的，把它放在坑里，换新坑
        a[pivot] = a[end];
        pivot = end;
        //左边找大的，如果不是大于key,继续
        while (begin < end && a[begin] <= key)
        {
            begin++;
        }
        //找到比key大的，把它放在坑里，换新坑
        a[pivot] = a[begin];
        pivot = begin;
    }

    a[pivot] = key;//bengin 与 end 相遇，相遇的位置一定是一个坑
    QuickSort1(a, left, pivot - 1);
    QuickSort1(a, pivot + 1, right);
}
int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
    QuickSort1(nums,0,numsSize-1);
    *returnSize = 0;
    *returnColumnSizes = (int *)malloc(sizeof(int)*100001);
    int *path = (int *)malloc(sizeof(int)*numsSize);
    int **result = (int **)malloc(sizeof(int *)*100001);
    int *visited = (int *)calloc(numsSize,sizeof(int));
    int *pathSize = (int *)calloc(1,sizeof(int));
    back(nums,numsSize,returnSize,returnColumnSizes,path,pathSize,result,visited);
    return result;

}

2、C++

class Solution {
public:
    vector<int> path;
    vector<vector<int>> result;
    void back(vector<int>& nums,vector<bool>& visited){
        if(path.size()==nums.size()){
            result.push_back(path);
            return;
        }
        for(int i=0;i<nums.size();i++){
            if(visited[i]||(i>0 && nums[i-1]==nums[i]&&visited[i-1]==false)){
                continue;
            }
            path.push_back(nums[i]);
            visited[i] = true;
            back(nums,visited);
            visited[i] = false;
            path.pop_back();
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<bool> visited(nums.size(),false);
        back(nums,visited);
        return result;
    }
};

3、JAVA

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    void back(int[] nums,boolean[] visited){
        if(nums.length == path.size()){
            result.add(new ArrayList<>(path));
            return;
        }
        for(int i=0;i<nums.length;i++){
            if(visited[i]){continue;}
            if(i>0&&nums[i-1]==nums[i]&&visited[i]==false&&visited[i-1]==false){
                continue;
            }
                visited[i] = true;
                path.add(nums[i]);
                back(nums,visited);
                path.removeLast();
                visited[i] = false;
        }
    }
    public List<List<Integer>> permuteUnique(int[] nums) {
        Arrays.sort(nums);
        boolean visited[] = new boolean[nums.length];
        Arrays.fill(visited,false);
        back(nums,visited);
        return result;
    }
}

4、Python

class Solution(object):
    def __init__(self):
        self.path = []
        self.result = []
    def back(self,nums,visited):
        if len(nums)==len(self.path):
            self.result.append(self.path[:])
            return
        for i in range(len(nums)):
            if visited[i]==True:
                continue
            if i>0 and nums[i-1]==nums[i] and visited[i-1]==False:
                continue
            visited[i]=True
            self.path.append(nums[i])
            self.back(nums,visited)
            self.path.pop()
            visited[i]=False

    def permuteUnique(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        visited = [False]*len(nums)
        self.back(nums,visited)
        return self.result