[Leetcode] Next Permutation

最新推荐文章于 2021-03-10 11:29:57 发布

原创最新推荐文章于 2021-03-10 11:29:57 发布 · 276 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

67 篇文章

订阅专栏

本文介绍了一个C++函数，用于将一组整数排列成字典序中下一个更大的排列。如果不存在这样的排列，则将其调整为最小排列。该函数通过在输入数组中找到合适的元素进行交换来实现这一目标。

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.

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

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

        int previous = -1;
        int current = -1;

        for(int i=0;i<num.size()-1;i++)
        {
            if(num[i]<num[i+1])
            {
                previous = i;
            }
        }

        if(previous!=-1)
        {
            for(int j=previous;j<num.size();j++)
            {
                if(num[previous]<num[j])
                {
                    current = j;
                }
            }
            swap(num[previous],num[current]);
            sort(num.begin()+previous+1,num.end());//注意sort函数的用法
        }
        else
        {
            sort(num.begin(),num.end());//lowest possible order
        }
    }
};