Leetcode: 31. 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.

Example

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

思路

• 从前往后找到最后一组前面数比后面数大的，这两个数不一定相邻
• 然后交换
• 交换完了之后，从交换完后的第一个数后面开始排序
• STL里面有个next_permutation(list, list)一步到位。。

  bool next_permutation( iterator start, iterator end);
  

代码

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int len = nums.size();
        if(len <= 1) return;
        
        int index1 = -1, index2 = -1, min_flag = -1;
        for(int i = 1; i < len; ++i){
            if(nums[i] == nums[i - 1]) continue;
            else if(nums[i] > nums[i - 1]){
                index1 = i - 1;
                index2 = i;
                min_flag = nums[i] - nums[i - 1];
            }
            else if(index1 != -1 && nums[i] > nums[index1] && nums[i] - nums[index1] <= min_flag ){
                index2 = i;
                min_flag = nums[i] - nums[index1];
            }
        }
        
        if(index1 != -1 && index2 != -1){
            swap(nums[index1], nums[index2]);
            sort(nums.begin() + index1 + 1, nums.end());
        }
        else sort(nums.begin(), nums.end());
    }
};