#52 Next Permutation

題目描述：

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]

題目思路：

這題注意到，正常的規律從後到前應該是後一個數比前一個數大。如果發現前一個數更小時，這就是發生轉折點的地方。首先，我們先找到這個轉折點（如果找不到，那麼答案就是reversed array）；然後從轉折點往後看，找出大於轉折點的最小數，並把這個數和轉折點互換。對於新轉折點後那些剩下的數，做一個sort。這樣，改完的數組就是答案了。

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;
    }
};