Leetcode: 1424. Diagonal Traverse II

Description

• 题目见 https://leetcode.com/problems/diagonal-traverse-ii/

code

这是考验智商的题目～～


我的代码

import bisect

class Solution(object):
    def findDiagonalOrder(self, nums):
        """
        :type nums: List[List[int]]
        :rtype: List[int]
        """
        
        ret = []
        N = len(nums)
        rows = [i for i in range(N)]
        
        def iter1(level):
            pos = bisect.bisect_left(rows, level)
            
            if level == N or pos >= len(rows) or rows[pos] != level:
                return
                      
            for j in reversed(rows[:pos+1]):
                if j > level:
                    continue
                p = level-j    
                if len(nums[j]) > p:
                    ret.append(nums[j][p])
                else:
                    rows.remove(j)                
            return iter1(level+1)
        
        def iter2(level, pos):
            if 0 > level or level not in rows:
                return
            if pos >= len(nums[level]):
                rows.remove(level)
                if len(rows) == 0:
                    return
                return iter2(rows[-1], pos+level-rows[-1])
            
            
            for j in reversed(rows):
                p = level-j+pos
                if len(nums[j]) > p:
                    ret.append(nums[j][p])
                else:
                    rows.remove(j)
                
            return iter2(level, pos+1)
        
        iter1(0)
        iter2(N-1, 1)
        return ret
          

lee215 代码
     def findDiagonalOrder(self, A):
        res = []
        for i, r in enumerate(A):
            for j, a in enumerate(r):
                if len(res) <= i + j:
                    res.append([])
                res[i + j].append(a)
        return [a for r in res for a in reversed(r)]

性能、简洁度都被吊打～