Next Permutation

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分类专栏： Math 文章标签： leetcode

于 2014-12-10 16:21:20 首次发布

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/41847633

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本文介绍了一个简单算法，用于求解整数数组的下一个字典序排列。如果不存在更大的排列，则将数组重新排列为最小的字典序排列。算法分为四步：找到最后一个升序对的位置，确定可交换的最大值，交换值并反转子数组。

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

思路：According to Wikipedia, a man named Narayana Pandita presented the following simple algorithm to solve this problem in the 14th century.

Find the largest index k such that nums[i] < nums[i + 1]. If no such index exists, just reverse nums and done.
Find the largest index j > i such that nums[i] < nums[j].
Swap nums[i] and nums[j].
Reverse the sub-array nums[i + 1 ~ n - 1].

class Solution {
    public void nextPermutation(int[] nums) {
        int n = nums.length;
        int i = n - 2;
        for(; i >= 0; i--) {
            if(nums[i] < nums[i + 1]) {
                break;
            }
        }
        
        if(i < 0) {
            reverse(nums, 0, n - 1);
            return;
        }
        int j = n - 1;
        while(j > i) {
            if(nums[j] > nums[i]) {
                break;
            } else {
                j--;
            }
        }
        swap(nums, i, j);
        reverse(nums, i + 1, n - 1);
    }
    
    private void swap(int[] A, int i, int j) {
        int temp = A[i];
        A[i] = A[j];
        A[j] = temp;
    }
    
    private void reverse(int[] A, int start, int end) {
        while(start <= end) {
            swap(A, start, end);
            start++;
            end--;
        }
    }
}