LintCode_190 Next Permutation II

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).

Example

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

Challenge 

The replacement must be in-place, do not allocate extra memory.

若果都是递减的， 那么已经到最大。所以只要找到那个不是递减的点， 然后找到比他大的最小的店， 交换， 然后把剩余布冯升序排列， 组合便可。

class Solution {public:    /**     * @param nums: a vector of integers     * @return: return nothing (void), do not return anything, modify nums in-place instead     */    void nextPermutation(vector<int> &nums) {        // write your code here        // write your code here        bool finished = false;        if (nums.size() <= 1)             finished = true;        priority_queue<int, vector<int>> pq;        vector<int> back_vec;        pq.push(INT_MIN);        unordered_map<int, int> map;        for (int i = nums.size() - 1; i >= 0 && !finished; i--) {             if (nums[i] < pq.top()) {                 back_vec.push_back(nums[i]);                 sort(back_vec.begin(), back_vec.end());                 auto itr = upper_bound(back_vec.begin(), back_vec.end(), nums[i]);                 nums[i] = *itr;                 back_vec.erase(itr);                 nums.erase(nums.begin() + i + 1, nums.end());                 nums.insert(nums.end(), back_vec.begin(), back_vec.end());                 finished = true;             } else {                 pq.push(nums[i]);                 back_vec.push_back(nums[i]);             }        }        if (!finished)           sort(nums.begin(), nums.end());    }};