Next Permutation问题及解法

问题描述：

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.

示例：

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

问题分析：

本题题意理解起来可能比较难，其实主要意思是求解一个数组的下一个全排列，详细过程可参考一下http://blog.csdn.net/qq575787460/article/details/41215475文章的讲解。

过程详见代码：

class Solution {public:    void nextPermutation(vector<int>& nums) {        if(nums.empty()) return;        int len = nums.size();        int i = len - 1;        for(;i > 0; i--)        {        if(nums[i] > nums[i-1])        {        for(int j = len - 1; j >= i; j--)        {        if(nums[j] > nums[i-1])        {        swap(nums[i-1],nums[j]);        break;}        }        sort(nums.begin() + i,nums.end());        break;}}if(i==0) sort(nums.begin(),nums.end());    }};