[Leetcode 31, medium] Next Permutation

Problem:

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

Solutions:

The solution is a typical back-tracking solution.

1. The definition of next permutation is a permutation of an array such that a'[i]=

Programs:

void swap(int &a, int &b){
        int temp=a;
        a=b;
        b=temp;
    }

    void nextPermutation(vector<int> &num) {
        if(num.size()<=1) return;
        
        int len=num.size();
        for(int i=len-1; i>=0; --i){
            if(i==0)
                for(int j=i, l=len-1; j<l; ++j, --l)    swap(num[j], num[l]);
            if(num[i-1]<num[i]){
                //Find the next value.
                for(int j=len-1; j>=i; --j){
                    if(num[j]>num[i-1]){
                        swap(num[i-1], num[j]);
                        break;
                    }
                }
                //Swap others.
                for(int j=i, l=len-1; j<l; ++j, --l)    swap(num[j], num[l]);
                break;
            }
        }
    }