leetcode_31. Next Permutation

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https://leetcode.com/problems/next-permutation/#/description

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

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題意：給你一個數組，問他的下一個排列是什麼。。。

思路：看幾組例子就可以知道了~~

1 2 3 5 4   ------>  1 2 4 3 5

1 2 5 3 4   ------>  1 3 2 4 5

1 3 2 5 4   ------>  1 3 4 2 5

所以，規律就是從後面往前找，找到一個nums[i-1]<nums[i]，然後再去後面找一個最小的比nums[i-1]大的，再對i~end的位置進行排序~~詳情見代碼

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int length = nums.size();
        int pos=-1;
        for(int i=length-1;i>0;i--)
        {
            if(nums[i-1]<nums[i])
            {
                pos=i-1;
                break;
            }
        }
        if(pos==-1)
        {
            sort(nums.begin(),nums.end());
            return;
        }
        int Min=1<<30;
        int p=0;
        for(int i=pos+1;i<length;i++)
        {
            if(Min>nums[i]&&nums[i]>nums[pos])
            {
                Min=nums[i];
                p=i;
            }
        }
        int t=nums[p];
        nums[p]=nums[pos];
        nums[pos]=t;
        sort(nums.begin()+pos+1,nums.end());
    }
};