leetcode——Next Permutation

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题目：

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

方法：

对于按照字典序进行求解，有一套方法：

1、依次将前后两个相邻数进行比较；找到后面大于前面的数的下标的最大值maxI；

2、从这个maxI开始（包括maxI），向后找到大于下标maxI-1上的数的下标的最大值maxL；

3、将L上的值跟maxI-1上的值互换，然后从maxI开始到末尾的子串进行颠倒即得到字典序的下一个序列。

AC代码：

public class Next_Permutation {    private void reverse(int[] num, int start, int end) {        while (start < end) {            int i = num[start];            num[start] = num[end];            num[end] = i;            start++;            end--;        }    }    public void nextPermutation(int[] num) {        if (num == null || num.length == 0)            return;        int maxI = -1, maxL = -1;        for (int i = 0; i < num.length - 1; i++) {            if (num[i] < num[i + 1]) {                maxI = i + 1;            }        }        if (maxI == -1) {            reverse(num, 0, num.length - 1);        } else {            for (int l = maxI; l < num.length; l++) {                if (num[l] > num[maxI - 1]) {                    maxL = l;                }            }            int i = num[maxI - 1];            num[maxI - 1] = num[maxL];            num[maxL] = i;            reverse(num, maxI, num.length - 1);        }    }}

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