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Next Permutation[leetcode]

基本思路是从后往前找到第一个递减的数num[index],并与之前的第一个大于num[index]的数交换位置。

注意交换后[index+1...n-1]仍是非递减的,所以只需要reverse一下,使其变成非递增的

    void nextPermutation(vector<int> &num) {
        int index = num.size() - 2;
        int rIndex = index;
        //traverse forward till find first decending num, save in index
        for (; index >= 0; index--)
        {
            //find first num larger than nun[index], save in rIndex
            if (num[index + 1] > num[index])
            {
                for (rIndex = num.size() - 1; rIndex > index && num[index] >= num[rIndex]; rIndex--);
                swap(*(num.begin() + index), *(num.begin() + rIndex));
                break;
            }
        }
        reverse(num.begin() + index + 1, num.end());//do not need sort, just reverse
    }


Next Permutation[leetcode]