LeetCode31. Next Permutation最清晰的解法_which rearranges numbers into the lexicographicall-优快云博客

本文介绍了一个用于重新排列数字数组以获得下一个更大排列的算法。该算法可以在原地完成操作，无需额外内存。通过一系列步骤，包括查找枢轴、找到其后续元素、交换以及反转后缀，实现从现有排列到下一个可能排列的转换。

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

解答：

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int pivot = findPivot(nums);
        int sidx = findPivotSuccessor(nums, pivot);
        if (pivot >= 0) {
            std::swap(nums[pivot], nums[sidx]);
        }
        reverseSuffix(nums, pivot+1);
    }
    
    int findPivot(vector<int>& nums)
    {
        for (int i = nums.size()-2; i >= 0; i--) {
            if (nums[i] < nums[i+1]) {
                return i;
            }
        }
        return -1;
    }
    
    int findPivotSuccessor(vector<int>&nums, int pivot)
    {
        int pv = nums[pivot];
        int idx = -1;
        for (int i = pivot+1; i < nums.size(); i++) {
            if (nums[i] > pv) {
                if (idx == -1) {
                    idx = i;
                } else if (nums[idx] > nums[i]) {
                    idx = i;
                }
            }
        }
        return idx;
    }
    
    void reverseSuffix(vector<int>& nums, int start)
    {
        std::sort(nums.begin()+start, nums.end());
    }
};