#52 Next Permutation

题目描述：

Given a list of integers, which denote a permutation.

Find the next permutation in ascending order.

 Notice

The list may contains duplicate integers.

Have you met this question in a real interview? 

Yes

Example

For [1,3,2,3], the next permutation is [1,3,3,2]

For [4,3,2,1], the next permutation is [1,2,3,4]

题目思路：

这题注意到，正常的规律从后到前应该是后一个数比前一个数大。如果发现前一个数更小时，这就是发生转折点的地方。首先，我们先找到这个转折点（如果找不到，那么答案就是reversed array）；然后从转折点往后看，找出大于转折点的最小数，并把这个数和转折点互换。对于新转折点后那些剩下的数，做一个sort。这样，改完的数组就是答案了。

Mycode（AC = 32ms）：

class Solution {public:    /**     * @param nums: An array of integers     * @return: An array of integers that's next permuation     */    vector<int> nextPermutation(vector<int> &nums) {        // write your code here        if (nums.size() <= 1) return nums;                // find the key position        int idx = -1;        for (int i = nums.size() - 2; i >= 0; i--) {            if (nums[i] < nums[i + 1]) {                idx = i;                break;            }        }                // if is the larget sequence, return reverse        if (idx == -1) {            reverse(nums.begin(), nums.end());            return nums;        }                int second_min = nums[idx + 1], second_idx = idx + 1;        for (int i = idx + 1; i < nums.size(); i++) {            if (nums[i] > nums[idx] && nums[i] < second_min) {                second_min = nums[i];                second_idx = i;            }        }                // swap        swap(nums, idx, second_idx);                // sort the rest numbers        sort(nums.begin() + idx + 1, nums.end());                return nums;    }        void swap(vector<int> &nums, int i, int j) {        int tmp = nums[i];        nums[i] = nums[j];        nums[j] = tmp;    }};