[leetcode]排列类问题

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本文详细介绍了NextPermutation算法的实现方法，该算法用于找出数组中下一个字典序更大的排列。若不存在这样的排列，则将其调整为最小的排列（即升序）。文章提供了具体的步骤和示例，帮助读者理解算法的工作原理。

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31. 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

Algorithm from wikipedia:
Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
Find the largest index l greater than k such that a[k] < a[l].
Swap the value of a[k] with that of a[l].
Reverse the sequence from a[k + 1] up to and including the final element a[n].

void nextPermutation(vector<int>& nums)
{
    int n = nums.size();
    if (n <= 1)
        return;
    int i;
    // get the rightmost that nums[i] < nums[i + 1]
    for (i = nums.size() - 2; i >= 0 && nums[i] >= nums[i + 1]; --i);
    if (i == -1)
    {
        reverse(nums.begin(), nums.end());
        return;
    }
    int j;
    // get the rightmost nums[j] that nums[j] > nums[i]
    for (j = i + 1; j < nums.size() && nums[i] < nums[j]; ++j);
    swap(nums[i], nums[--j]);
    // reverse from i + 1 to the end
    reverse(nums.begin() + i + 1, nums.end());
}