LeetCode(31): Next Permutation (C++)

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本文介绍如何实现将一组数字重新排列为字典序中下一个更大的排列。若无法形成更大的排列，则将其排列为最小的顺序（升序）。提供两种解决方案：一种利用标准库函数next_permutation()，另一种手动实现算法。

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

方案一

直接偷懒，用 std::next_permatation()函数即可实现。

void nextPermutation(vector<int>& nums) {
    next_permutation(nums.begin(), nums.end());
}

方案二

动手实现，算法流程如下图所示

算法描述如下

1. From right to left, find the first element (Partition Element) which violate the 
   increase trend, in this example, 5 will be selected since 8, 7, 4, 3, 2 already 
   in a increase trend.
2. From right to left, find the first element which large than Partition Element, 
   call it Change Element. Here 7 will be selected.
3. Swap the Partition Element and Change Element.
4. Reverse all the elements on the rigth of Partition Element.

代码如下

class Solution {
private:
    template<typename BidiIter>
    bool nextPermutation(BidiIter first, BidiIter last){
        const auto rfirst = reverse_iterator<BidiIter>(last);
        const auto rlast = reverse_iterator<BidiIter>(first);
    
        auto pivot = next(rfirst);
    
        while(pivot != rlast && *pivot >= *prev(pivot)){
            ++pivot;
        }
    
        if(pivot == rlast){
            reverse(rfirst, rlast);
            return false;
        }
    
        auto change = find_if(rfirst, pivot, bind1st(less<int>(), *pivot));
    
        swap(*change, *pivot);
        reverse(rfirst, pivot);
    
        return true;
    }
public:
    void nextPermutation(vector<int>& nums) {
        nextPermutation(nums.begin(), nums.end());
    }
};