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 and use only constant 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 → 1,3,23,2,1 → 1,2,31,1,5 → 1,5,1
思路：
1. 从后向前找两个相邻的元素，前一个位置记为i，后一个位置记为j，满足nums[i] < nums[j]
2. 从后向前找另一个位置k，满足nums[i] < nums[k]， 将nums[i]与nums[j]交换，将j（包含j）之后所有元素逆序，得到下一个排序数
class Solution { public: void nextPermutation(vector<int>& nums) { if (nums.empty()) return; int n = nums.size(); int i = 0, j = n-1, k = n-1; while (j > 0) { if (nums[j] > nums[j-1]) { i = j - 1; break; } j--; } while (k > i) { if (nums[k] > nums[i]) { swap(nums[k], nums[i]); break; } k--; } reverse(nums.begin() + j, nums.end()); } };
原文：https://www.cnblogs.com/immjc/p/9340839.html