[LeetCode]31. Next Permutation

最新推荐文章于 2020-12-07 13:42:30 发布

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最新推荐文章于 2020-12-07 13:42:30 发布

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分类专栏： LeetCode 文章标签： leetcode

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本文介绍了一个LeetCode上的经典问题“下一个排列”的解决方案。该问题要求在原地修改输入数组，使其变为字典序中下一个更大的排列。若不存在这样的排列，则将其重新排列为最小的可能顺序（即升序）。文章详细解释了解决方案的思路，并提供了完整的Java代码实现。

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Problem Description

[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

思路

找规律的题，多写几个就能找到规律了。

Code

package q031;

public class Solution {

    public void nextPermutation(int[] nums) {
        int n = nums.length;
        if (n <= 1)
            return;
        int i;
        for (i = n - 1; i >= 1; i--) {
            if (nums[i] > nums[i - 1])
                break;
        }
        if (i == 0) {
            reverse(nums, 0, n - 1);
        } else {
            int val = nums[i - 1];
            int j = n - 1;
            while (j >= i) {
                if (nums[j] > val)
                    break;
                j--;
            }
            swap(nums, j, i - 1);
            reverse(nums, i, n - 1);
        }

    }

    public void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
        // nums[i] = nums[i] ^ nums[j];
        // nums[j] = nums[i] ^ nums[j];
        // nums[i] = nums[i] ^ nums[j];
    }

    public void reverse(int[] nums, int s, int e) {
        if (s > e)
            return;
        for (int i = s; i <= (s + e) / 2; i++)
            swap(nums, i, s + e - i);
    }

}