[LeetCode]46.Permutations

【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].

【分析】

无

【代码】

/*********************************
*   日期：2015-01-16
*   作者：SJF0115
*   题目: 46.Permutations
*   网址：https://oj.leetcode.com/problems/permutations/
*   结果：AC
*   来源：LeetCode
*   博客：
**********************************/
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > result;
        if(num.empty()){
            return result;
        }//if
        vector<int> visited;
        // 递归
        DFS(num,visited,result);
        return result;
    }
private:
    void DFS(vector<int> &num,vector<int> &visited,vector<vector<int> > &result){
        // 形成一个全排列
        if(num.size() == visited.size()){
            result.push_back(visited);
            return;
        }//if
        vector<int>::iterator isVisited;
        for(int i = 0;i < num.size();i++){
            // 判断num[i]是否已经访问过
           isVisited = find(visited.begin(),visited.end(),num[i]);
           // 如果没有访问过
           if(isVisited == visited.end()){
               visited.push_back(num[i]);
               DFS(num,visited,result);
               visited.pop_back();
           }//if
        }//for
    }
};

int main(){
    Solution solution;
    vector<int> num;
    num.push_back(1);
    num.push_back(2);
    num.push_back(3);
    num.push_back(4);
    // 重新排列
    vector<vector<int> > permutes = solution.permute(num);
    // 输出
    for(int i = 0;i < permutes.size();i++){
        cout<<"[";
        for(int j = 0;j < permutes[i].size();j++){
            cout<<permutes[i][j];
        }//for
        cout<<"]"<<endl;
    }
    return 0;
}

【分析二】

直接调用STL中的next_permutation函数，如果在面试中面试官肯定会让你自己实现next_permutation函数

【代码二】

/*********************************
*   日期：2015-01-17
*   作者：SJF0115
*   题目: 46.Permutations
*   网址：https://oj.leetcode.com/problems/permutations/
*   结果：AC
*   来源：LeetCode
*   时间复杂度：O(n!)
*   空间复杂度：O(1)
*   博客：
**********************************/
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > result;
        // 排序
        sort(num.begin(),num.end());
        // 直接调用STL中的next_permutation
        do{
            result.push_back(num);
        }while(next_permutation(num.begin(),num.end()));
        return result;
    }
};

int main(){
    Solution solution;
    vector<int> num;
    num.push_back(1);
    num.push_back(2);
    num.push_back(3);
    num.push_back(4);
    // 重新排列
    vector<vector<int> > permutes = solution.permute(num);
    // 输出
    for(int i = 0;i < permutes.size();i++){
        cout<<"[";
        for(int j = 0;j < permutes[i].size();j++){
            cout<<permutes[i][j];
        }//for
        cout<<"]"<<endl;
    }
    return 0;
}

【分析三】

自己实现next_permutation函数

【代码三】

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > result;
        // 排序
        sort(num.begin(),num.end());
        vector<int> tmp = num;
        // 全排列的下一排列
        result.push_back(num);
        nextPermutation(num);
        while(num!= tmp){
            result.push_back(num);
            nextPermutation(num);
        }//while
        return result;
    }
private:
    void nextPermutation(vector<int> &num) {
        int count = num.size();
        if(count == 0 || count == 1){
            return;
        }//if
        // From right to left,find first digit which violate the increase
        // call it partitionNumber
        int index;
        for(index = count-2;index >= 0;index--){
            if(num[index+1] > num[index]){
                break;
            }//if
        }//for
        int tmp;
        // -1 indicate largest
        if(index != -1){
            // From right to left,find first digit which large than partitionNumber
            // call it changeNumber
            for(int i = count-1;i > index;i--){
                if(num[index] < num[i]){
                    // swap the partition index and changeNumber
                    tmp = num[index];
                    num[index] = num[i];
                    num[i] = tmp;
                    break;
                }//if
            }//for
        }//if
        // reverse all the digit on the right of partition index
        for(int i = index+1,j = count - 1;i < j;i++,j--){
            tmp = num[i];
            num[i] = num[j];
            num[j] = tmp;
        }//for
    }
};