954. 二倍数对数组

English Version

题目描述

给定一个长度为偶数的整数数组 arr，只有对 arr 进行重组后可以满足 “对于每个 0 <= i < len(arr) / 2，都有 arr[2 * i + 1] = 2 * arr[2 * i]” 时，返回 true；否则，返回 false。

 

示例 1：

输入：arr = [3,1,3,6]
输出：false

示例 2：

输入：arr = [2,1,2,6]
输出：false

示例 3：

输入：arr = [4,-2,2,-4]
输出：true
解释：可以用 [-2,-4] 和 [2,4] 这两组组成 [-2,-4,2,4] 或是 [2,4,-2,-4]

 

提示：

• 0 <= arr.length <= 3 * 104
• arr.length 是偶数
• -105 <= arr[i] <= 105

解法

方法一：哈希表 + 排序

Python3

class Solution:
    def canReorderDoubled(self, arr: List[int]) -> bool:
        freq = Counter(arr)
        if freq[0] & 1:
            return False
        for x in sorted(freq, key=abs):
            if freq[x << 1] < freq[x]:
                return False
            freq[x << 1] -= freq[x]
        return True

Java

class Solution {
    public boolean canReorderDoubled(int[] arr) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int v : arr) {
            freq.put(v, freq.getOrDefault(v, 0) + 1);
        }
        if ((freq.getOrDefault(0, 0) & 1) != 0) {
            return false;
        }
        List<Integer> keys = new ArrayList<>(freq.keySet());
        keys.sort(Comparator.comparingInt(Math::abs));
        for (int k : keys) {
            if (freq.getOrDefault(k << 1, 0) < freq.get(k)) {
                return false;
            }
            freq.put(k << 1, freq.getOrDefault(k << 1, 0) - freq.get(k));
        }
        return true;
    }
}

C++

class Solution {
public:
    bool canReorderDoubled(vector<int>& arr) {
        unordered_map<int, int> freq;
        for (int& v : arr) ++freq[v];
        if (freq[0] & 1) return false;
        vector<int> keys;
        for (auto& [k, _] : freq) keys.push_back(k);
        sort(keys.begin(), keys.end(), [](int a, int b) { return abs(a) < abs(b); });
        for (int& k : keys) {
            if (freq[k * 2] < freq[k]) return false;
            freq[k * 2] -= freq[k];
        }
        return true;
    }
};

Go

func canReorderDoubled(arr []int) bool {
	freq := make(map[int]int)
	for _, v := range arr {
		freq[v]++
	}
	if freq[0]%2 != 0 {
		return false
	}
	var keys []int
	for k := range freq {
		keys = append(keys, k)
	}
	sort.Slice(keys, func(i, j int) bool {
		return abs(keys[i]) < abs(keys[j])
	})
	for _, k := range keys {
		if freq[k*2] < freq[k] {
			return false
		}
		freq[k*2] -= freq[k]
	}
	return true
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

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