[Leetcode] Next permutation

Next permutation

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If Such arrangement is not a possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must is in-place, do not allocate extra memory.

Here is some examples. Inputs is in the left-hand column and its corresponding outputs is in the right-hand column.
1,2,3→1,3,2
3,2,1→1,2,3
1,1,5→1,5,1

Problem Solving Ideas:

This is an algorithmic problem. Learned a new way. Happy. The solution is as follows:

Class Solution {Public:void nextpermutation (vector<int> &num) {int len = num.size ();        if (len<2) {return;        }//int target1=-1 from right to left in the first reverse order;                for (int i=len-2; i>=0; i--) {if (num[i]<num[i+1]) {target1=i;            Break            }} if (target1>=0) {int target2=-1;                    for (int i=len-1; i>target1; i--) {if (Num[i]>num[target1]) {target2=i;                Break        }} swap (num, target1, Target2);    } reverse (num, target1+1, len-1);        } void Swap (vector<int>& num, int i, int j) {int temp=num[i];        NUM[I]=NUM[J];    Num[j]=temp;            } void reverse (vector<int>& num, int i, int j) {while (i<j) {swap (num, I, j);            i++;        j--; }    }};

[Leetcode] Next permutation