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
思路：从后向前找，找到第一个不是逆序的位置，也就是找到一个数，它大于它后面的那个数。然后将它后面的那个数开始到数组结尾的逆序都变为正序，并且从中找出第一个大于找到的那个数并交换二者的位置。
C++代码实现：
#include<iostream> #include<vector> using namespace std; class Solution { public: void nextPermutation(vector<int> &num) { if(num.empty()||num.size()==1) return; int i,j,k; int len=num.size(); for(i=int(num.size()-1); i>0; i--) { if(num[i]<=num[i-1]) continue; else break; } if(i==0) { i=0; k=len-1; while(i<k) { swap(&num[i],&num[k]); i++; k--; } return; } j=i-1; k=len-1; while(i<k) { swap(&num[i],&num[k]); i++; k--; } for(i=j+1; i<len; i++) if(num[i]>num[j]) break; swap(&num[i],&num[j]); } void swap(int *a,int *b) { int temp=*a; *a=*b; *b=temp; } }; int main() { Solution s; vector<int> vec= {3,2,1,4,5,9,8,7}; s.nextPermutation(vec); for(auto a:vec) cout<<a<<" "; cout<<endl; }
运行结果：
原文：http://www.cnblogs.com/wuchanming/p/4113060.html