LeetCode周赛#114 Q2 Array of Doubled Pairs

题目来源：https://leetcode.com/contest/weekly-contest-114/problems/array-of-doubled-pairs/

问题描述

 954. Array of Doubled Pairs

Given an array of integers A with even length, return true if and only if it is possible to reorder it such that A[2 * i + 1] = 2 * A[2 * i]for every 0 <= i < len(A) / 2.

 

Example 1:

Input: [3,1,3,6]

Output: false

Example 2:

Input: [2,1,2,6]

Output: false

Example 3:

Input: [4,-2,2,-4]

Output: true

Explanation: We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4].

Example 4:

Input: [1,2,4,16,8,4]

Output: false

 

Note:

1. 0 <= A.length <= 30000
2. A.length is even
3. -100000 <= A[i] <= 100000

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题意

给定一个序列，判断序列能否按某种顺序重新排列，使得奇数位的元素是前面一个偶数位元素的两倍

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思路

将序列排序，按负数序列、0序列、正数序列分别判断。

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代码

class Solution {
public:
    bool canReorderDoubled(vector<int>& A) {
        sort(A.begin(), A.end());
        int n = A.size(), i, j, neg = -1, pos = -1, zero = 0;   
        // neg: position of last negative element, pos: position of first positive element, zero: # of zeros
        if (n == 0)
        {
            return true;
        }
        vector<int> vis(n, 0);
        for (i=0; i<n; i++)
        {
            if (A[i] < 0)
            {
                neg = i;
            }
            else if (A[i] > 0)
            {
                pos = i;
                break;
            }
            else
            {
                zero++;
            }
        }
        if (zero % 2 != 0 || (neg != -1 && neg % 2 == 0) || (pos != -1 && (n-pos) % 2 == 1))
        {
            return false;
        }
        if (neg != -1)
        {
            for (i=neg; i>0; i--)
            {
                if (vis[i] == 0)
                {
                    vis[i] = 1;
                    for (j=i-1; j>=0; j--)
                    {
                        if (vis[j] == 0 && A[j] == 2*A[i])
                        {
                            vis[j] = 1;
                            break;
                        }
                    }
                    if (j == -1)
                    {
                        return false;
                    }
                }
            }
        }
        if (pos != -1)
        {
            for (i=pos; i<n-1; i++)
            {
                if (vis[i] == 0)
                {
                    vis[i] = 1;
                    for (j=i+1; j<n; j++)
                    {
                        if (vis[j] == 0 && A[j] == 2*A[i])
                        {
                            vis[j] = 1;
                            break;
                        }
                    }
                    if (j == n)
                    {
                        return false;
                    }
                }
            }
        }
        return true;
    }
};