LeetCode 31. Next Permutation

最新推荐文章于 2020-10-03 00:35:45 发布

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最新推荐文章于 2020-10-03 00:35:45 发布

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原文链接：http://www.cnblogs.com/gavinxing/p/5532720.html

本文提供了一个关于LeetCode上‘下一个排列’问题的有效解决方案。该算法从数组末尾开始向前查找，找到第一个可以与后面较大数值交换的位置，并将之后的数组进行升序排序，实现寻找下一个字典序排列。

Problem:

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

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

Thought:

　　from end to begin, find the first number that have number greater than it after, then swap the number with the minimum greater number, the sort the array after the number.

　　e.g 2 6 3 4 3 1 find arr[2] = 3, the swap it with arr[3] = 4, sort arr[3] to arr[5]

Code C++:

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        if (nums.size() <= 1)
            return;
        
        for (int i = nums.size() - 2; i >= 0; i--) {
            int greater_min = i;//greater_min point to the minimum number greater than nums[i]
            
            for (int j = nums.size() - 1; j > i; j--) {//get greater_min
                if (nums[j] > nums[i]) {
                    if (greater_min == i) {
                        greater_min = j;
                        continue;
                    }
                    greater_min  = nums[j] < nums[greater_min] ? j : greater_min;
                }
            }
            
            if (greater_min != i) {
                swap(nums[i], nums[greater_min]);
                sort(nums.begin() + i + 1, nums.end());
                return;
            }
        }
        sort(nums.begin(), nums.end());
        return;
    }
};