算法作业HW23：LeetCode 31 Next Permutation

Description:

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

 

Note:

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

Solution:

  Analysis and Thinking:

题目要求对于给定的输入序列，给出其字典序更大的下一个序列，这里用了简单的几个循环，从后向前扫描，不断找到升序的相邻元素，然后进行交与反转，从而得到下一个字典排序

  Steps:

1.从后向前扫描，找到第一对升序排列的相邻元素

2.如果i<0，表示已经找到最大排列为当前排列，其下一排列就是最小排列，直接反转

3.如果i>0，说明找到子序列，从当前序列后向前扫描，找到第一个比i下标元素大的j下标元素

4.交换i，j下标，从j开始反转

Codes:

class Solution {public:    void nextPermutation(vector<int> &num) {        if (num.size() < 2) return;        int i, j;        for (i = num.size() - 2; i >= 0; --i)             if (num[i+1] > num[i])                 break;        if (i < 0)        {            reverse(num.begin() , num.end());            return;        }        for (j = num.size() - 1; j > i && i >= 0 ; --j)            if (num[j] > num[i])                 break;        if (i >= 0)        {            swap(num[j], num[i]);            reverse(num.begin() + i + 1, num.end());            return;        }    }};

Results: