31. Next Permutation (Array; Math)

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).

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

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

思路：对大小的影响尽可能小=>影响尽可能右侧的位=>右向左扫描，找到第一个<右侧的数(i)

将右边最小的>nums[i]的数(j)与它交换

从小到大排列i之后的数

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int size = nums.size();
        int tmp, i ,j;

        for(i = size-; i >= ; i--){
            for(j = i+; j < size; j++){
                if(nums[i] >= nums[j]) continue;

                //meet the first num < at least one number at right
                for(int k = j+; k < size; k++){ //find the smallest one > nums[i]
                    if(nums[i] >= nums[k] || nums[j] <= nums[k]) continue;

                    //find the smaller one > nums[i]
                    j = k;
                }
                tmp = nums[i];
                nums[i]=nums[j];
                nums[j]=tmp;
                sort(nums.begin()+i+, nums.end());
                break;
            }
            if(j<size) break;
        }
        if(i<){
            sort(nums.begin(), nums.end());
        }
    }
};

To make codes more simple, we can integrate k iterates into j iterates

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int size = nums.size();
        int swapIndex = size, tmp, i ,j;

        for(i = size-; i >= ; i--){
            for(j = i+; j < size; j++){
                if(nums[j] <= nums[i] || (swapIndex<size && nums[j]>=nums[swapIndex])) continue;
                swapIndex = j;
            }
            if(swapIndex>=size) continue;
            tmp = nums[i];
            nums[i]=nums[swapIndex];
            nums[swapIndex]=tmp;
            sort(nums.begin()+i+, nums.end());
            break;
        }
        if(i<){
            sort(nums.begin(), nums.end());
        }
    }
};