870 Advantage Shuffle 优势洗牌
Description:
You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].
Return any permutation of nums1 that maximizes its advantage with respect to nums2.
Example:
Example 1:
Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]
Example 2:
Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]
Constraints:
1 <= nums1.length <= 10^5
nums2.length == nums1.length
0 <= nums1[i], nums2[i] <= 10^9
题目描述:
给定两个大小相等的数组 A 和 B，A 相对于 B 的优势可以用满足 A[i] > B[i] 的索引 i 的数目来描述。
返回 A 的任意排列，使其相对于 B 的优势最大化。
示例 :
示例 1：
输入：A = [2,7,11,15], B = [1,10,4,11]
输出：[2,11,7,15]
示例 2：
输入：A = [12,24,8,32], B = [13,25,32,11]
输出：[24,32,8,12]
提示:
1 <= A.length = B.length <= 10000
0 <= A[i] <= 10^9
0 <= B[i] <= 10^9
思路:
贪心
对 A 排序
将 B 中元素和对应下标插入优先队列或者排序
每次选择一个 A 中最小大于当前数字的数
如果 A 中最大的数都比当前数小, 则选最小的
时间复杂度为 O(nlgn), 空间复杂度为 O(n)
代码:
C++:
class Solution
{
public:
vector advantageCount(vector& nums1, vector& nums2)
{
int n = nums1.size(), left = 0, right = n - 1, right2 = n;
vector result(n, 0);
vector> v;
for (int i = 0; i < n; i++) v.emplace_back(make_pair(nums2[i], i));
sort(nums1.begin(), nums1.end());
sort(v.begin(), v.end());
while (--right2 > -1)
{
if (nums1[right] > v[right2].first) result[v[right2].second] = nums1[right--];
else result[v[right2].second] = nums1[left++];
}
return result;
}
};
Java:
class Solution {
public int[] advantageCount(int[] nums1, int[] nums2) {
int n = nums1.length, left = 0, right = n - 1, result[] = new int[n];
Arrays.sort(nums1);
PriorityQueue pq = new PriorityQueue<>((a, b) -> { return b[1] - a[1]; });
for (int i = 0; i < n; i++) pq.offer(new int[] { i, nums2[i] });
while (!pq.isEmpty()) {
int[] cur = pq.poll();
if (nums1[right] > cur[1]) result[cur[0]] = nums1[right--];
else result[cur[0]] = nums1[left++];
}
return result;
}
}
Python:
class Solution:
def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
sorted1, sorted2, assigned, remaining, i = sorted(nums1), sorted(nums2), { num: [] for num in nums2 }, [], 0
for num in sorted1:
if num > sorted2[i]:
assigned[sorted2[i]].append(num)
i += 1
else:
remaining.append(num)
return [assigned[num].pop() if assigned[num] else remaining.pop() for num in nums2]