[LeetCode] (medium) 31. Next Permutation

https://leetcode.com/problems/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 and use only constant 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

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字典序的下一个排列等价于相同前缀尽量长，所以从后往前寻找能够“变大”的最短后缀，也就是后缀中的第一个数不是当前后缀中的最大值。

class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int cur = nums.size()-2;
        while(cur >= 0){
            if(nums[cur] >= nums[cur+1]) --cur;
            else break;
        }
        if(cur < 0){
            sort(nums.begin(), nums.end());
        }else{
            int tem = binary_search(nums, nums[cur], cur, nums.size()-1);
            swap(nums[cur], nums[tem]);
            sort(nums.begin()+(cur+1), nums.end());
        }
    }
    
    int binary_search(vector<int> &nums, int target, int lo, int hi){
        int mid;
        while(lo < hi){
            mid = (lo+hi+1)/2;
            if(nums[mid] > target) lo = mid;
            else hi = mid-1;
        }
        return lo;
    }
};