leetcode 46-Permutations and 47-Permutations II

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].

Permutations II

 

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

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

全排列问题

第二个题目基于第一题。第二题给的数组中还有可能相等的元素。

请參照 全 排 列

题一代码

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
       vector<vector<int>>v;
       v.clear();
       vector<int>a;
       next_c(v,nums.size(),a,nums,0);
       return v;
    }
    void next_c(vector<vector<int>>&v,int n,vector<int>& a,vector<int>& b,int cur) /// b[]中的数能够同样
    {
        if(cur==n){
            v.push_back(a);
        }
        else{
            for(int i=0;i<n;i++)
                 if(!i||b[i]!=b[i-1])///
                 {
                    int c1=0,c2=0;
                    for(int j=0;j<cur;j++) if(a[j]==b[i]) c1++;
                    for(int j=0;j<n;j++) if(b[j]==b[i]) c2++;
                    if(c1<c2){
                        a.push_back(b[i]);// 不能用a[cur]=b[i];
                        next_c(v,n,a,b,cur+1);
                        a.pop_back();
                      }
                 }
        }
    }
};

题二代码。当然能够过题一。

class Solution {
public:
    vector<vector<int>>v;
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        v.clear();
       vector<int>a;a.clear();
       sort(nums.begin(),nums.end());
       next_c(nums.size(),a,nums,0);
       return v;
    }
    void next_c(int n,vector<int>& a,vector<int>& b,int cur) /// b[]中的数能够同样
{
    if(cur==n){
        v.push_back(a);
    }
    else{
        for(int i=0;i<n;i++)
        if(!i||b[i]!=b[i-1])///
            {
            int c1=0,c2=0;
            for(int j=0;j<cur;j++) if(a[j]==b[i]) c1++;
            for(int j=0;j<n;j++) if(b[j]==b[i]) c2++;
            if(c1<c2){
                a.push_back(b[i]); //a[cur]=b[i];
                next_c(n,a,b,cur+1);
                a.pop_back();
            }
        }
    }
}
};