题目链接
难度： 中等 类型：贪心算法
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]
解题思路
排序A和B, 如果A中最小的数大于B中最小的数，则将A的最小数分配给B的最小数。 否则A放入remain列表中。其余的数以此类推分析。
代码实现
class Solution(object):
def advantageCount(self, A, B):
sortedA = sorted(A)
sortedB = sorted(B)
# assigned[b] = list of a that are assigned to beat b
# remaining = list of a that are not assigned to any b
assigned = {b: [] for b in B}
remaining = []
# populate (assigned, remaining) appropriately
# sortedB[j] is always the smallest unassigned element in B
j = 0
for a in sortedA:
if a > sortedB[j]:
assigned[sortedB[j]].append(a)
j += 1
else:
remaining.append(a)
# Reconstruct the answer from annotations (assigned, remaining)
return [assigned[b].pop() if assigned[b] else remaining.pop()
for b in B]
