【LeetCode】46. Permutations (2 solutions)

排列算法详解

最新推荐文章于 2024-03-31 10:58:20 发布

转载最新推荐文章于 2024-03-31 10:58:20 发布 · 91 阅读

本文详细介绍了两种求解全排列问题的方法。第一种为递归法，通过不断交换元素位置来生成所有可能的排列组合。第二种方法则利用了next_permutation函数，先对数组进行排序，然后迭代获取所有升序排列。

Permutations

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].

解法一：递归

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > ret;
        Helper(ret, num, 0);
        return ret;
    }
    void Helper(vector<vector<int> >& ret, vector<int> num, int pos)
    {
        if(pos == num.size()-1)
            ret.push_back(num);
        else
        {
            for(int i = pos; i < num.size(); i ++)
            {//swap all the ints to the current position
                swap(num[pos], num[i]);
                Helper(ret, num, pos+1);
                swap(num[pos], num[i]);
            }
        }
    }
};

解法二：just a joke

别忘了先排序，因为next_permutation是升序返回的。

class Solution 
{
public:
    vector<vector<int> > permute(vector<int> &num) 
    {
        vector<vector<int> > result;
        //sort first
        //note that next_permutation is in ascending order
        sort(num.begin(), num.end());
        result.push_back(num);
        while(next_permutation(num.begin(), num.end()))
        {
            result.push_back(num);
        }
        return result;
    }
};