LeetCode31. Next Permutation（C++）

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 and use only constant 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

求下一个排列，STL中有next_permutation函数emmmm

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

好了，不开玩笑，这题的意思就是要自己实现下一个排列。

要实现下一个排列，不妨想一想，既然是下一个，肯定不能拿高位开刀，我们从后面往前遍历，找到一个不满足递增的，下一个的排列一定是从已经遍历的数中挑出一个比这个数大，但是最小的数与之交换，那剩下的低位满足递增排列即可

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