LeetCode 31 - Next Permutation

最新推荐文章于 2020-12-07 13:42:30 发布

EnjoyCodingAndGame

最新推荐文章于 2020-12-07 13:42:30 发布

阅读量324

点赞数

CC 4.0 BY-SA版权

分类专栏：编程题目文章标签： C++ LeetCode

本文链接：https://blog.youkuaiyun.com/ywcpig/article/details/50739142

编程题目专栏收录该内容

78 篇文章

订阅专栏

本文介绍了一个名为NextPermutation的算法实现，该算法用于将一组数字重新排列为字典序中下一个更大的排列。若无法形成更大排列，则将其重排为最小可能的顺序。算法采用原地替换方式，不分配额外内存。

Next Permutation

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

My Code

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int size = nums.size();
        if (size < 1)
            return;

        int max = nums.back();

        int i = size - 2;
        for (; i >= 0; i--)
        {
            if (nums[i] < max)
            {
                int greater = max + 1;
                int idx = -1;
                for (int j = i + 1; j < size; j++)
                    if (nums[j] > nums[i] && nums[j] < greater)
                    {
                        greater = nums[j];
                        idx = j;
                    }

                nums[idx] = nums[i];
                nums[i] = greater;

                sort(nums.begin()+i+1, nums.end());

                break;
            }
            else
                max = nums[i];
        }

        if (i == -1)
            sort(nums.begin(), nums.end());
    }
};

Runtime: 12 ms