2017/11/20 Leetcode 日记

2017/11/14 Leetcode 日记

442. Find All Duplicates in an Array

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        int sz = nums.size();
        int once[sz+2];
        vector<int> num;
        for(int i = 0; i < sz+2; i++){
            once[i] = 0;
        }
        for(int i = 0; i < sz; i++){
            if(once[nums[i]] == 1){
                num.push_back(nums[i]);
            }else{
                once[nums[i]] ++;
            }
        }
        return num;
    }
};

class Solution:
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        num = []
        for x in nums:
            if nums[abs(x)-1] < 0:
                num.append(abs(x))
            else:
                nums[abs(x)-1] *= -1
        return num

406. Queue Reconstruction by Height

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

(先以第一序列升序，第二序列降序排列，再插入表中。)

vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
    auto comp = [](const pair<int, int>& p1, const pair<int, int>& p2)
                    { return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second); };
    sort(people.begin(), people.end(), comp);
    vector<pair<int, int>> res;
    for (auto& p : people) 
        res.insert(res.begin() + p.second, p);
    return res;
}