114th LeetCode Weekly Contest 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

题目很短，题意很明显，说一下思路吧

首先得分出正数和负数，0就不管了都是0，正数是小到大，负数则反过来，然后我们只找两个数字一组，为什么是两个数字呢

如果是2 4 4 8，把2,4,8找了，那还有个4怎么办，所以应该是2,4   4,8就可以了

操作就是拿出现的数字*2去看有没有这个数，最后是否还留下数字

class Solution {
public:
    bool canReorderDoubled(vector<int>& A) {
        sort(A.begin(),A.end());
        int len = A.size();
        map<int,int>Mp;
        vector<int>Ve1;
        vector<int>Ve2;
        vector<int>Ve;
        set<int>Se;
        int num;
        for(int i=0;i<len;i++){
            Mp[A[i]]++;
            if(A[i]>0){
                Ve1.push_back(A[i]);
            }
            if(A[i]<0){
                Ve2.push_back(A[i]);
            }
        }
        int len1 = Ve1.size();
        for(int i=0;i<len1;i++){
            Ve.clear();
            num = Ve1[i];
            while(Mp[num]>0){
                Mp[num]--;
                Ve.push_back(num);
                num*=2;
                if(Ve.size()==2){
                    break;
                }
            }
            for(int j=0;j<Ve.size();j++){
                //cout<<Ve[j]<<" ";
            }
            //cout<<endl;
            if(Ve.size()==1){
                Mp[Ve1[i]]++;
            }
        }
        int len2 = Ve2.size();
        for(int i=len2-1;i>=0;i--){
            Ve.clear();
            num = Ve2[i];
            while(Mp[num]>0){
                Mp[num]--;
                Ve.push_back(num);
                num*=2;
                if(Ve.size()==2){
                    break;
                }
            }
            for(int j=0;j<Ve.size();j++){
                //cout<<Ve[j]<<" ";
            }
            //cout<<endl;
            if(Ve.size()==1){
                Mp[Ve2[i]]++;
            }
        }
        for(int i=0;i<len;i++){
            //cout<<Mp[A[i]]<<"A"<<endl;
            if(Mp[A[i]]&&A[i]!=0){
                return false;
            }
        }
        return true;
    }
};