leetcode Next Permutation-优快云博客

本文链接：https://blog.youkuaiyun.com/shashafbqc/article/details/30030645

本文介绍了一种在原地生成下一个字典序全排列的算法。该算法首先找到当前排列中可以进行交换以生成更大排列的最大元素位置，然后与后续元素中大于该位置元素的最大值进行交换，并将剩余部分反转。

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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 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

组合数学P32 用字典序法生成下一个全排列

class Solution {
public:
/*组合数学，下一个全排列的生成。考虑排列P，求满足Pj-1<pj的j的最大值，设为i
求满足关系式Pi-1<pk的k 的最大值，设为j。将pi-1与pj交换，并将pi(包括pi)之后的元素逆序,因为pi（包括pi）之后的元素现在是倒序*/
    void nextPermutation(vector<int> &num) {
        int n=num.size();
        int i=n-2;
        while(i>=0){
            if(num[i]>=num[i+1])
                i--;
            else
                break;
        }
        if(i<0){
            reverse(num,0,n-1);
            return;
        }
        int k=n-1;
        while(k>=0&&num[k]<=num[i]){
            k--;
        }
        swap(num[i],num[k]);
        reverse(num,i+1,n-1);
    
        
    }
    void  swap(int &a,int &b){
        int temp=a;
        a=b;
        b=temp;
    }
    void reverse(vector<int> &num,int start,int end){
        while(start<end){
            swap(num[start++],num[end--]);
        }
    }
};