描述
You are given a 0-indexed array nums consisting of positive integers. You can choose two indices i and j, such that i != j, and the sum of digits of the number nums[i] is equal to that of nums[j].
Return the maximum value of nums[i] + nums[j] that you can obtain over all possible indices i and j that satisfy the conditions.
Example 1:
Input: nums = [18,43,36,13,7]
Output: 54
Explanation: The pairs (i, j) that satisfy the conditions are:
- (0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54.
- (1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50.
So the maximum sum that we can obtain is 54.
Example 2:
Input: nums = [10,12,19,14]
Output: -1
Explanation: There are no two numbers that satisfy the conditions, so we return -1.
Note:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
解析
根据题意，给定一个由正整数组成的索引为 0 的数组 nums 。 您可以选择两个索引 i 和 j ，使得 i != j，并且数字 nums[i] 的位数之和等于 nums[j] 的位数之和。返回 nums[i] + nums[j] 的最大值，可以在满足条件的所有可能索引 i 和 j 上获得该最大值。
这道题其实很简单，我们定义字典 d ，d 的键位数字各位数之和，d 的值为这种键对应出现的数字 n ，然后我们只需要遍历 d 中的键值对 k ，v ，如果 v 的长度大于 1， 那么我们就将 v 进行排序，找出最大的两个值之和 v[-1] + v[-2] 去和 result 进行比较取最大值更新 result 即可。遍历结束之后返回 result 即可。
时间复杂度为 O(NlogN) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def maximumSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = -1
d = collections.defaultdict(list)
for n in nums:
d[self.digitSum(n)].append(n)
for k,v in d.items():
if len(v) > 1:
v.sort()
result = max(result, v[-1] + v[-2])
return result
def digitSum(self, n):
if n == 0:
return 0
return self.digitSum(n//10) + n % 10
运行结果
82 / 82 test cases passed.
Status: Accepted
Runtime: 1464 ms
Memory Usage: 26.5 MB
原题链接
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