topic：

2679.In the matrix and the harmony.md

Thought：

One -line First of all, the meaning is to find the largest number of each sub -list，Thenpopgo out，Finally ask for peace。 The effect of traversing again and again is too bad，So I thought of using itzip一次性Traversal多个子Array。 于yes先对每个子ArraySort，Then usezipTraversal，Find the maximum。 for example： nums = [[7,2,1],[6,4,2],[6,5,3],[3,2,1]] Sort后： nums = [[1,2,7],[2,4,6],[3,5,6],[1,2,3]] Then usezipTraversal得到： [(1,2,3,1),(2,4,5,2),(7,6,6,3)] Find the maximum： [3,5,7] Context： 15

Code：

class Solution:
    def matrixSum(self, nums: List[List[int]]) -> int:
        return sum(max(i) for i in \
        zip(*(sorted(sublist) for sublist in nums)))

*The role and the rolezip()explain

nums = [[1,2,7],[2,4,6],[3,5,6],[1,2,3]]

for i in range(len(nums[1])):
    for j in range(len(nums)):
        print(nums[j][i])
# ans = 123124527663
num1 = [1,2,7]
num2 = [2,4,6]
num3 = [3,5,6]
num4 = [1,2,3]

zip()The function corresponds to one -to -one elements in multiple lists，Then return onezipObject，Can uselist()Function convert to list。

for i in zip(num1, num2, num3, num4):
    print(i)
#(1, 2, 3, 1)
#(2, 4, 5, 2)
#(7, 6, 6, 3)

*numsThe role ispythonNot a pointer，InsteadnumsEach element in the parameter is passed into the function。I understand here as a list。

# Enumerate
print(*nums)
# [1, 2, 7] [2, 4, 6] [3, 5, 6] [1, 2, 3]

zip(*nums)WillnumsEach element is passed in as a parameterzip()In the function，Then return onezipObject，Can uselist()Function convert to list。 zip(*nums)Equivalent tozip(num1, num2, num3, num4)，innum1, num2, num3, num4yesnumsElement。

for i in zip(*nums):
    print(i)
# Equivalent
#(1, 2, 3, 1)
#(2, 4, 5, 2)
#(7, 6, 6, 3)

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