【LeetCode】031. Next Permutation

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

题解：

　　from here

Solution 1 ()

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int i, j, n = num.size();
        for (i = n - 2; i >= 0; --i) {
            if (num[i + 1] > num[i]) {
                for (j = n - 1; j >= i; --j) {
                    if (num[j] > num[i]) break;
                }
                swap(num[i], num[j]);
                reverse(num.begin() + i + 1, num.end());
                return;
            }
        }
        reverse(num.begin(), num.end());
    }
};

　　Solution 2-4 are from here

Solution 2 ()

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int i = nums.size() - 1, k = i;
        while (i > 0 && nums[i-1] >= nums[i])
            i--;
        for (int j=i; j<k; j++, k--)
            swap(nums[j], nums[k]);
        if (i > 0) {
            k = i--;
            while (nums[k] <= nums[i])
                k++;
            swap(nums[i], nums[k]);
        }
    }
};

　　使用STL库函数

Solution 3 ()

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        auto i = is_sorted_until(nums.rbegin(), nums.rend());
        if (i != nums.rend())
            swap(*i, *upper_bound(nums.rbegin(), i, *i));
        reverse(nums.rbegin(), i);
    }
};　

　　　使用STL库函数

Solution 4 ()

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