[LeetCode] (medium) 31. Next Permutation

最新推荐文章于 2021-05-21 19:06:52 发布

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最新推荐文章于 2021-05-21 19:06:52 发布

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分类专栏： LeetCode 文章标签： LeetCode

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实现一个函数，用于生成给定数字序列的字典序下一个更大的排列。如果无法生成更大的排列，则将数字重新排列为最小的排列（即升序排列）。替换必须在原地进行，并且仅使用常量额外内存。

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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

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;
    }
};