Permutations I(II) 数组的全排列

题目

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

思路：递归

class Solution {public:    vector<vector<int> > permute(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > result;        myfun(result,num,0);        return result;    }            void myfun(vector<vector<int> > &result, vector<int> &num, int k)    {        if(k==num.size()-1)            result.push_back(num);        else        {            for(int i=k;i<num.size();i++)            {                int tmp = num[i];                num[i] = num[k];                num[k] = tmp;                myfun(result,num,k+1);                tmp = num[i];                num[i] = num[k];                num[k] = tmp;            }        }          }     };

如果元素有重复，则需要加入判断才可以，往下递归。

class Solution {public:    vector<vector<int> > permuteUnique(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > result;        myfun(result,num,0);        return result;    }            void myfun(vector<vector<int> > &result, vector<int> &num, int k)    {        if(k==num.size()-1)            result.push_back(num);        else        {            for(int i=k;i<num.size();i++)            {                if(isunique(num,k,i))                {                             int tmp = num[i];                    num[i] = num[k];                    num[k] = tmp;                    myfun(result,num,k+1);                    tmp = num[i];                    num[i] = num[k];                    num[k] = tmp;                }            }        }          }         bool isunique(vector<int> &num, int k,int i)    {        for(int j=k;j<i;j++)            if(num[j]==num[i])                return false;        return true;    }};

 

最新 java

public class Solution {    public List<List<Integer>> permuteUnique(int[] nums) {        List<List<Integer>> result = new ArrayList<List<Integer>>();         if(nums.length == 0){            return result;        }        permute(nums, 0, result);        return result;    }    public void permute(int[] nums, int cur, List<List<Integer>> result) {        if(cur >= nums.length){            List<Integer> list = convertArrayToList(nums);            result.add(list);        }        for(int i=cur; i<nums.length; i++){            if(!isuniq(nums, cur, i)){                continue;            }            int temp = nums[cur];            nums[cur] = nums[i];            nums[i] = temp;            permute(nums, cur+1, result);            temp = nums[cur];            nums[cur] = nums[i];            nums[i] = temp;        }    }    private ArrayList<Integer> convertArrayToList(int[] num) {    ArrayList<Integer> item = new ArrayList<Integer>();    for (int h = 0; h < num.length; h++) {    item.add(num[h]);    }    return item;    }    private boolean isuniq(int[] nums, int from, int to){        for(int i=from; i<to; i++){            if(nums[i] == nums[to]){                return false;            }        }        return true;    }}

我也试过其他的思路，但是都失败了。

class Solution {public:    vector<vector<int> > permute(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > result;        bool flag = true;        result.push_back(num);                int i=0;        for(int i=1;i<num.size()-1;i++)        {            for(int j=0;j<num.size()-i;j++)            {                int tmp = num[j];                num[j] = num[j+i];                num[j+i] = tmp;                result.push_back(num);                tmp = num[j];                num[j] = num[j+i];                num[j+i] = tmp;            }                      }        return result;            }};

上述迭代遍历的代码对于

[1,2,3] 返回结果是[[1,2,3],[2,1,3],[2,3,1],[1,3,2]] 而真实结果是[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]改进还在继续。