题目描述
爱丽丝和鲍勃拥有不同总数量的糖果。给你两个数组 aliceSizes
和 bobSizes
,aliceSizes[i]
是爱丽丝拥有的第 i
盒糖果中的糖果数量,bobSizes[j]
是鲍勃拥有的第 j
盒糖果中的糖果数量。
两人想要互相交换一盒糖果,这样在交换之后,他们就可以拥有相同总数量的糖果。一个人拥有的糖果总数量是他们每盒糖果数量的总和。
返回一个整数数组 answer
,其中 answer[0]
是爱丽丝必须交换的糖果盒中的糖果的数目,answer[1]
是鲍勃必须交换的糖果盒中的糖果的数目。如果存在多个答案,你可以返回其中 任何一个 。题目测试用例保证存在与输入对应的答案。
示例 1:
输入:aliceSizes = [1,1], bobSizes = [2,2]
输出:[1,2]
示例 2:
输入:aliceSizes = [1,2], bobSizes = [2,3]
输出:[1,2]
示例 3:
输入:aliceSizes = [2], bobSizes = [1,3]
输出:[2,3]
示例 4:
输入:aliceSizes = [1,2,5], bobSizes = [2,4]
输出:[5,4]
提示:
1 <= aliceSizes.length, bobSizes.length <= 104
1 <= aliceSizes[i], bobSizes[j] <= 105
- 爱丽丝和鲍勃的糖果总数量不同。
- 题目数据保证对于给定的输入至少存在一个有效答案。
解法
方法一
| class Solution:
def fairCandySwap(self, aliceSizes: List[int], bobSizes: List[int]) -> List[int]:
diff = (sum(aliceSizes) - sum(bobSizes)) >> 1
s = set(bobSizes)
for a in aliceSizes:
target = a - diff
if target in s:
return [a, target]
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21 | class Solution {
public int[] fairCandySwap(int[] aliceSizes, int[] bobSizes) {
int s1 = 0, s2 = 0;
Set<Integer> s = new HashSet<>();
for (int a : aliceSizes) {
s1 += a;
}
for (int b : bobSizes) {
s.add(b);
s2 += b;
}
int diff = (s1 - s2) >> 1;
for (int a : aliceSizes) {
int target = a - diff;
if (s.contains(target)) {
return new int[] {a, target};
}
}
return null;
}
}
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18 | class Solution {
public:
vector<int> fairCandySwap(vector<int>& aliceSizes, vector<int>& bobSizes) {
int s1 = accumulate(aliceSizes.begin(), aliceSizes.end(), 0);
int s2 = accumulate(bobSizes.begin(), bobSizes.end(), 0);
int diff = (s1 - s2) >> 1;
unordered_set<int> s(bobSizes.begin(), bobSizes.end());
vector<int> ans;
for (int& a : aliceSizes) {
int target = a - diff;
if (s.count(target)) {
ans = vector<int>{a, target};
break;
}
}
return ans;
}
};
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| function fairCandySwap(aliceSizes: number[], bobSizes: number[]): number[] {
let s1 = aliceSizes.reduce((a, c) => a + c, 0);
let s2 = bobSizes.reduce((a, c) => a + c, 0);
let diff = (s1 - s2) >> 1;
for (let num of aliceSizes) {
let target = num - diff;
if (bobSizes.includes(target)) {
return [num, target];
}
}
}
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