LeetCode Next Permutation

最新推荐文章于 2021-03-10 11:29:57 发布

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本文详细介绍了如何使用迭代器和交换技巧实现下一个排列算法，确保序列转换为字典序中下一个更大的排列。若无法找到更大排列，则将其排列为最小可能顺序。

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

class Solution {
public:
    void nextPermutation(vector<int> &num) {
    	int n = num.size();
        if(n <= 1)
        	return;
        //从尾部开始找第一个前面小于后面的数 
        int i;
        for(i = n-1; i >=1; i--) {
        	if(num[i-1] < num[i])
        		break;
        }
        int pre = i-1;
        int cur = i;
        //该序列已经是最后一个，返回第一个 
        if(cur == 0) {
        	sort(num.begin(), num.end());
        	return;
        }
        //从尾部找第一个大于num[pre]的数 
        int j;
        for(j = n-1; j >= cur; j--) {
        	if(num[j] > num[pre])
        		break;
        }
        swap(num[pre], num[j]);
        sort(num.begin()+ cur, num.end());    
    }
};