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LeetCode 31. Next Permutation

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题目

原题在此

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

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

The replacement must be in place and use only constant extra memory.

Example 1:
Input: nums = [1,2,3]
Output: [1,3,2]

Example 2:
Input: nums = [3,2,1]
Output: [1,2,3]

Example 3:
Input: nums = [1,1,5]
Output: [1,5,1]

Example 4:
Input: nums = [1]
Output: [1]

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 100

解析

有std::next_permutation可用, 一行搞定.
或者, 根据wiki有如下方法:

  1. 找到最大下标k满足a[k] < a[k+1]. 若没有则已经是最后一个置换.
  2. 找到最大下边l满足l > k && a[l] > a[k].
  3. swap(a[k], a[l]).
  4. 调换a[k+1]至结尾的所有元素.

代码

std

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class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        next_permutation(begin(nums), end(nums));
    }
};

impl

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class Solution {
public:
    void nextPermutation(vector<int>& nums) {
    	int n = nums.size(), k, l;
    	for (k = n - 2; k >= 0; k--) {
            if (nums[k] < nums[k + 1]) {
                break;
            }
        }
    	if (k < 0) {
    	    reverse(nums.begin(), nums.end());
    	} else {
    	    for (l = n - 1; l > k; l--) {
                if (nums[l] > nums[k]) {
                    break;
                }
            } 
    	    swap(nums[k], nums[l]);
    	    reverse(nums.begin() + k + 1, nums.end());
        }
    }
};