leetcode 31. 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

是数学中全排列

思路

根据规则，从后往前找到第一个逆序点，然后从这个点开始后，找到最小的大于逆序点小值的数，二者兑换后，对后面的整体进行反转。
即a0a1a2...aN-1的数下，找到 a[i]<a[i+1] && a[i+1]>a[i+2]>....>a[N-1]。
然后找到a[j]>a[i] && a[j+1]<=a[i], 对调i，j后，a[i+1,....,N-1]形成非升序，翻转为非降序即可。

参考：

http://blog.csdn.net/nomasp/article/details/49913627

https://segmentfault.com/a/1190000007285442

代码

    public void nextPermutation(int[] nums) {    if (nums.length == 1) {    //如果长度为1，直接返回    return;    }        int index = nums.length - 1;    while ( (index != 0) && (nums[index-1] >= nums[index])) {    //从后找到第一个逆序点    index--;    }    index--;    if (index == -1) {      //如果是654321这样的，直接全部倒序    Arrays.sort(nums);    return;    }    int min = Integer.MAX_VALUE;    int minindex = index;    for (int i = index + 1 ; i < nums.length ; i++) {    //在逆序点之后找到第一个比它小的数    if (nums[i] > nums[index] && nums[i] < min) {    min = nums[i];    minindex = i;    }    }    int tmp = nums[index];    //交换    nums[index] = nums[minindex];    nums[minindex] = tmp;    //    int last = (int) Math.floor((index + nums.length)/2);//    int j = nums.length - 1;//    for (int i = (index + 1) ; i <= last ; i++) {//    int temp = nums[i];//    nums[i] = nums[j];//    nums[j] = temp;//    j--;//    }    Arrays.sort(nums,index+1,nums.length);   //逆序点之后排序，而不是完全交换    }

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