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English Version

题目描述

给定两个整数数组,请交换一对数值(每个数组中取一个数值),使得两个数组所有元素的和相等。

返回一个数组,第一个元素是第一个数组中要交换的元素,第二个元素是第二个数组中要交换的元素。若有多个答案,返回任意一个均可。若无满足条件的数值,返回空数组。

示例:

输入: array1 = [4, 1, 2, 1, 1, 2], array2 = [3, 6, 3, 3]
输出: [1, 3]

示例:

输入: array1 = [1, 2, 3], array2 = [4, 5, 6]
输出: []

提示:

  • 1 <= array1.length, array2.length <= 100000

解法

先计算两个数组的差值 diff,若 diff 为奇数,则说明无满足条件的数值,返回空数组。否则,将 array2 转为 set。然后遍历 array1 中的每个数 a,若值 a - diffset 中,则说明找到满足条件的数值对。

Python3

class Solution:
    def findSwapValues(self, array1: List[int], array2: List[int]) -> List[int]:
        diff = sum(array1) - sum(array2)
        if diff & 1:
            return []
        diff >>= 1
        s = set(array2)
        for a in array1:
            b = a - diff
            if b in s:
                return [a, b]
        return []

Java

class Solution {
    public int[] findSwapValues(int[] array1, int[] array2) {
        int s1 = 0, s2 = 0;
        Set<Integer> s = new HashSet<>();
        for (int a : array1) {
            s1 += a;
        }
        for (int b : array2) {
            s.add(b);
            s2 += b;
        }
        int diff = s1 - s2;
        if ((diff & 1) == 1) {
            return new int[] {};
        }
        diff >>= 1;
        for (int a : array1) {
            int b = a - diff;
            if (s.contains(b)) {
                return new int[] {a, b};
            }
        }
        return new int[] {};
    }
}

C++

class Solution {
public:
    vector<int> findSwapValues(vector<int>& array1, vector<int>& array2) {
        int s1 = 0, s2 = 0;
        unordered_set<int> s;
        for (int a : array1) s1 += a;
        for (int b : array2) {
            s2 += b;
            s.insert(b);
        }
        int diff = s1 - s2;
        if (diff & 1) {
            return {};
        }
        diff >>= 1;
        for (int a : array1) {
            int b = a - diff;
            if (s.count(b)) {
                return {a, b};
            }
        }
        return {};
    }
};

Go

func findSwapValues(array1 []int, array2 []int) []int {
	s1, s2 := 0, 0
	for _, a := range array1 {
		s1 += a
	}
	s := make(map[int]bool)
	for _, b := range array2 {
		s2 += b
		s[b] = true
	}
	diff := s1 - s2
	if (diff & 1) == 1 {
		return []int{}
	}
	diff >>= 1
	for _, a := range array1 {
		b := a - diff
		if s[b] {
			return []int{a, b}
		}
	}
	return []int{}
}

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