常见算法问题之NextPermutation

题目来源于leetcode 31

---

1. 问题描述

依据字典序升序，找出序列的下一序列。例如（左边是输入，右边是输出）：

1 2 3 --- 1 3 21 5 1 --- 5 1 13 2 1 --- 1 2 3

---

2. 算法描述

for (i = 0; i < size(array); i++)    j = max{i | array[i] < array[i + 1]};for (i = 0; i < size(array); i++)    k = max{i | array[i] > array[j]};swap(j, k);reverse(array, j+1, array(end));

---

3. C++实现

#include <iostream>#include <vector>using namespace std;class Solution {public:    void nextPermutation(vector<int>& nums)     {        vector<int>::iterator i, j, k, reverse_start = nums.begin();        bool is_found = 0; //若原序列恰好是降序，则reverse范围是整个序列        for (i = nums.begin(), j = i; i != nums.end() - 1; i++)        {            if (*i < *(i + 1))            {                j = i;                is_found = 1;             }        }        if (is_found != 0)        {            for (i = nums.begin(); i != nums.end(); i++)            {                if (*j < *i)                    k = i;            }            reverse_start = j + 1;            swap(j, k);        }        reverse(nums, reverse_start);    }private:    void reverse(vector<int>& nums, vector<int>::iterator start)    {        vector<int>::iterator i = start;        vector<int>::iterator j = nums.end() - 1;        while (i < j)        {            swap(i, j);            i++;            j--;        }    }    void swap(vector<int>::iterator i, vector<int>::iterator j)    {        int temp = *i;        *i = *j;        *j = temp;    }};int main(){    const int n = 3;    int b[n] = {3, 2, 1};    vector<int> a(b, b + n);    Solution S;    S.nextPermutation(a);    vector<int>::iterator iter;    for (iter = a.begin(); iter != a.end(); iter++)    {        cout << *iter << ' ';    }    cout << endl;    return 0;}

---

参考：
1. 字典序：http://www.cnblogs.com/pmars/p/3458289.html
2. LeetCode原题：https://leetcode.com/problems/next-permutation/
3. C++ vector使用：http://blog.csdn.net/liunian17/article/details/7435781