[LeetCode]Next Permutation

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

class Solution {//step1：从右往左找到第一个破坏升序(非严格)的元素，此例中为4.记下标为 i//step2: 依然从右往左,找到第一个大于4的元素，此例中5，交换4和5.//step3：从i+1到最右端，逆置。public:    void nextPermutation(vector<int>& nums) {        int len = nums.size();        int temp = INT_MIN;        int index = -1; //无升序        for(int i=nums.size()-1; i>=0; --i){            if(nums[i]<temp){ //非严格升序                index = i;                break;            }            temp = nums[i];        }        for(int i=nums.size()-1; i>=0; --i){            if(index == -1) break; //严格升            if(nums[i]>nums[index]){                int temp = nums[i];                nums[i] = nums[index];                nums[index] = temp;                break;            }        }        swap(nums,index+1,len-1);    }    void swap(vector<int>& nums,int i,int j){        while(i<j){            int temp = nums[i];            nums[i++] = nums[j];            nums[j--] = temp;        }    }};