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  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 → 1,3,2

3,2,1 → 1,2,3

1,1,5 → 1,5,1

找数组全排列中，按字典序比较的下一个组合。

思路：找第一组顺序对(nums[i] > nums[i-1]). 在i之后找比nums[i-1]大的最小数nums[k]，然后把nums[k]换到nums[i-1]位置，最后对i开始后面所有的数做一次排序。

1 class Solution { 2 public: 3     void nextPermutation(vector
     
      & nums) { 4         int len = nums.size(); 5         for (int i=len-1; i>=1; --i) { 6             if (nums[i] > nums[i-1]) { 7                 int key = nums[i-1]; 8                 swap(nums[i], nums[i-1]); 9                 for (int j=len-1; j>i; --j) {10                     if ((nums[j] > key) && (nums[j] < nums[i-1]))11                         swap(nums[i-1], nums[j]);12                 }13                 sort(nums.begin()+i, nums.end());14                 return;15             }16         }17         sort(nums.begin(), nums.end());18         return;19     }20 };