31.Next Permutation

思路：

• 参考 http://blog.csdn.net/abcjennifer/article/details/40152323， 复杂度为 (O(nlog(n)))

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
    	int i,k = 0;
        for(i = nums.size()-1; i > 0; i--){
        	if(nums[i] > nums[i-1])  break;
        }
        k = i;
        if(k == 0) {
        	sort(nums.begin(), nums.end());
        }
        else{
        	for(i = nums.size()-1; i >= k; i--){
        		if(nums[i] > nums[k-1]) break;
        	}
        	swap(nums[i],nums[k-1]);
        	sort(nums.begin()+k, nums.end());
        }
        //for(k = 0; k < nums.size(); k++)
        //	cout << nums[k] << " ";
    }
};