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

分析：找下一個排列。排列的過程能夠理解爲：

一、從右至左找到第一個不知足遞增規律的數，例如 1 3 2 5 4，那麼2應該被找出，記下標爲left

二、從右至左找到第一個大於nums[left]的數，記下標爲right

三、將nums[left]與nums[right]交換

四、將left右邊的數字所有倒轉

運行時間 16ms

 1 class Solution {
 2 public:
 3     void nextPermutation(vector<int>& nums) {
 4         if(nums.size() == 1) return;
 5         if(nums.size() == 2){
 6             reverse(nums.begin(), nums.end());
 7             return;
 8         }
 9         
10         //find left
11         int left = -1, right = nums.size() - 1;
12         for(int i = nums.size() - 1; i >= 1; i--){
13             if(nums[i] > nums[i-1]){
14                 left = i - 1;
15                 break;
16             }
17         }
18         if(left == -1){
19             sort(nums.begin(), nums.end());
20             return;
21         }
22         else{
23             //find right
24             for(int j = nums.size() - 1; j > left; j--){
25                 if(nums[j] > nums[left]){
26                     right = j;
27                     break;
28                 }
29             }
30             //swap nums[left] and nums[right]
31             int temp = nums[left];
32             nums[left] = nums[right];
33             nums[right] = temp;
34             
35             //reverse the numbers after left
36             int indexLeft = left + 1, indexRight = nums.size() - 1;
37             while(indexLeft < indexRight){
38                 int temp1 = nums[indexLeft];
39                 nums[indexLeft++] = nums[indexRight];
40                 nums[indexRight--] = temp1;
41             }
42             return;
43         }
44     }
45 };