[Leetcode] Find Original Array From Doubled Array(Medium)
LeetCode 2007 - Find Original Array From Doubled Array
An integer array original is transformed into a doubled array changed by appending twice the value of every element in original, and then randomly shuffling the resulting array.
Given an array changed, return original if changed is a doubled array. If changed is not a doubled array, return an empty array. The elements in original may be returned in any order.
example
Input: changed = [1,3,4,2,6,8]
Output: [1,3,4]
Explanation: One possible original array could be [1,3,4]:
- Twice the value of 1 is 1 * 2 = 2.
- Twice the value of 3 is 3 * 2 = 6.
- Twice the value of 4 is 4 * 2 = 8.
Other original arrays could be [4,3,1] or [3,1,4].
Input: changed = [6,3,0,1]
Output: []
Explanation: changed is not a doubled array.
Input: changed = [1]
Output: []
Explanation: changed is not a doubled array.
How can we solve this problem?
這是其實需要我們計算每個數出現的頻率，如果要找出某個數x的雙倍數是否存在於array中，如果存在，我們只需要將x與雙倍數移除(因為每個x只會匹配到一個順便數)。如果不存在，也就代表了這個array並不是一個有效的Double list，直接return false即可。
Solution:
class Solution {
public:
vector<int> findOriginalArray(vector<int>& changed) {
if(changed.size() % 2 != 0) return {}; //must be an even size
vector<int> res;
int n = changed.size();
sort(changed.begin(),changed.end());
unordered_map<int,int> m;
for(auto num : changed) m[num]++; //number frequency
for(auto num : changed){
if(m.count(num) && m[num] > 0){ //num is exist?
m[num]--;
if(!m.count(num * 2) || m[num*2] == 0){ //num * 2 is not exist or num*2 is empty
return {};
}else {
m[num * 2]--;
res.push_back(num);
}
}
}
return res;
}
};