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

這一題的時間複雜度是O(n)。 解題思路：

從後往前找，m,j,i,l, ， 若是num[i] > num[i + 1], 即 一直是倒序，則繼續往前找， 直到找個一個num[i]< num[i + 1]的， 即第一個比它右邊小的數字。 

這是由於若是 從後往前一直是倒序，則已是permutation了 不可能出現更大的， 只有順序反了的那個地方纔是須要動的地方。

接着將這個數字num[i] , 和它右邊的數字比較，找個一個恰好比他大一點的數字， 二者替換位置。 接着再將i 以後的數字排序，正序，便可。

 1 public class Solution {
 2     public void nextPermutation(int[] num) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int j,i;
 6         for(i = num.length - 1; i > 0; i --){
 7             j = i - 1;
 8             if(num[j] < num[i]){
 9                 int ex = 0;
10                 int a;
11                 for(a = i; a < num.length; a ++){
12                     if(num[a] > num[j]){
13                         ex = num[a];
14                     }
15                     else{
16                         break;
17                     }
18                 }
19                 num[a - 1] = num[j];
20                 num[j] = ex;
21                 for(int ii = i; ii < (num.length - i) / 2 + i; ii ++){
22                     int cur = num[ii];
23                     num[ii] = num[num.length - ii - 1+ i];
24                     num[num.length - ii- 1 + i] = cur;
25                 }
26                 return;
27             }
28         }
29         for(i = 0; i < num.length / 2; i ++){
30             int cur = num[i];
31             num[i] = num[num.length - 1 - i];
32             num[num.length - 1 - i] = cur;
33         }
34     }
35 }

 

 第三遍：

 1 public class Solution {
 2     public void nextPermutation(int[] num) {
 3         if(num == null || num.length < 2) return;
 4         int len = num.length;
 5         int point = -1;
 6         for(point = len - 2; point > -1; point --){
 7             if(num[point] < num[point + 1]) break;
 8         }
 9         if(point == -1){
10             for(int i = 0; i < num.length / 2; i ++){
11                 int cur = num[i];
12                 num[i] = num[num.length - 1 - i];
13                 num[num.length - 1 - i] = cur;
14             }
15             return;
16         }
17         int pos = 0;
18         for(int i = point + 1; i < len; i ++){
19             if(num[i] > num[point]) pos = i;
20         }
21         int ttmp = num[point];
22         num[point] = num[pos];
23         num[pos] = ttmp; 
24         
25         for(int i = 1; i <= (len - point - 1) / 2; i ++){
26             int tmp = num[point + i];
27             num[point + i] = num[len - i];
28             num[len - i] = tmp;
29         }
30     }
31 }