[LeetCode] Next Permutation

hale1007

于 2014-05-03 15:56:27 发布

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分类专栏： LeetCode Algorithm

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本文介绍了一种用于生成数字序列下一个字典序排列的算法NextPermutation。该算法能在原地将一组数字重新排列为字典序上的下一个更大的排列。若无法生成更大的排列，则将数字重新排列为最小的排列（即升序排列）。文章提供了两种实现方式：一种是直接调用STL中的next_permutation函数，另一种是手动实现该算法。

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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

这个题目是实现STL算法next_permutation，如果只要过这个题目，直接调用即可，但也可以自己简易实现之

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        next_permutation(num.begin(), num.end());
    }
};

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        if(num.empty() || num.size() == 1) return;
        int first = 0, last = num.size();
        int i = last - 1;
        while(1)
        {
            int ii = i--;
            if(num[i] < num[ii])
            {
                int j = last;
                while(num[--j] <= num[i]);
                swap(num[i], num[j]);
                reverse(num.begin() + ii, num.end());
                return;
            }
            if(i == 0)
            {
                reverse(num.begin(), num.end());
                return;
            }
        }
    }
};