LeetCode 31.Next Permutation

最新推荐文章于 2020-12-07 13:42:30 发布

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最新推荐文章于 2020-12-07 13:42:30 发布

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本文介绍了一个C++实现的nextPermutation函数，该函数用于将一组数字重新排列为字典序中下一个更大的排列。若无法实现更大排列，则将其排列为最小可能的顺序（即升序）。本文提供了详细的代码实现和解析。

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题目：

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

分析解答：

class Solution {
public:
	void nextPermutation(vector<int> &num) {
		vector<int>::size_type n = num.size();
		if(n==0 || n==1){
		    return;
		}
		int swap_in = 0,swap_out = 0;
		size_t i = n - 1 , j = 0;
		for(i = n - 1; i != 0;--i){
			if(num[i-1] < num[i]){
				swap_in = i - 1;
				break;
			}
		}
		if(i != 0){
			for(i = n - 1; i != swap_in;--i){
				if(num[i] > num[swap_in]){
					swap_out = i;
					break;
				}
			}
			vector_swap(num,swap_in,swap_out);
			j = swap_in + 1;
		}else{
		    j = 0;
		}
		i = n - 1;
		while(j < i){
			vector_swap(num,j,i);
			++j;
			--i;
		}
	}
	void vector_swap(vector<int> &v,int m,int n){
		int temp = v[m];
		v[m] = v[n];
		v[n] = temp;
	}
};