题目：原题链接（中等）
标签：贪心算法
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 1 + N 2 ) O(N1+N2) O(N1+N2) | O ( 6 ) O(6) O(6) | 108ms (93.17%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
c1, c2 = collections.Counter(nums1), collections.Counter(nums2)
s1, s2 = sum(i * c1[i] for i in range(1, 7)), sum(i * c2[i] for i in range(1, 7))
if s1 == s2:
return 0
if s1 > s2:
s1, s2 = s2, s1
c1, c2 = c2, c1
# s1 < s2
diff = s2 - s1
ans = 0
if c1[1] > 0 or c2[6] > 0:
if (c1[1] + c2[6]) * 5 >= diff:
return ans + math.ceil(diff / 5)
else:
diff -= (c1[1] + c2[6]) * 5
ans += (c1[1] + c2[6])
if c1[2] > 0 or c2[5] > 0:
if (c1[2] + c2[5]) * 4 >= diff:
return ans + math.ceil(diff / 4)
else:
diff -= (c1[2] + c2[5]) * 4
ans += (c1[2] + c2[5])
if c1[3] > 0 or c2[4] > 0:
if (c1[3] + c2[4]) * 3 >= diff:
return ans + math.ceil(diff / 3)
else:
diff -= (c1[3] + c2[4]) * 3
ans += (c1[3] + c2[4])
if c1[4] > 0 or c2[3] > 0:
if (c1[4] + c2[3]) * 2 >= diff:
return ans + math.ceil(diff / 2)
else:
diff -= (c1[4] + c2[3]) * 2
ans += (c1[4] + c2[3])
if c1[5] > 0 or c2[2] > 0:
if (c1[5] + c2[2]) * 1 >= diff:
return ans + math.ceil(diff / 1)
else:
diff -= (c1[5] + c2[2]) * 1
ans += (c1[5] + c2[2])
return -1
