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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 → 1,3,23,2,1 → 1,2,31,1,5 → 1,5,1
以上方法中，最坏情况需要扫描数组三次，所以时间复杂度是O(3*n)=O(n)，空间复杂度是O(1)。代码如下：
1 class Solution { 2 public: 3 void nextPermutation(vector<int> &num) { 4 5 int n=num.size(); 6 int i,j; 7 for(i=n-2;i>=0;i--) 8 { 9 if(num[i]<num[i+1]) 10 { 11 break; 12 } 13 } 14 if(i>=0) 15 { 16 for(j=i+1;j<n;j++) 17 { 18 if(num[j]<=num[i]) 19 { 20 break; 21 } 22 } 23 j--; 24 swap(num[i],num[j]); 25 } 26 27 reverse(num.begin()+i+1,num.end()); 28 } 29 30 31 void swap(int &a,int &b) 32 { 33 int tmp=a; 34 a=b; 35 b=tmp; 36 } 37 };
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原文地址：http://www.cnblogs.com/reachteam/p/4214181.html
