@@ -652,6 +652,37 @@ class Solution:
652652 r = {'':r, '.':r, '..':r[:-1]}.get(s, r + [s])
653653 return '/' + '/'.join(r)
654654```
655+ 656+ ## [ 73. 矩阵置零 5行] ( https://leetcode-cn.com/problems/set-matrix-zeroes/ )
657+ ``` python
658+ class Solution :
659+ def setZeroes (self matrix : List[List[int ]]) -> None :
660+ row = [[0 ] * len (i) if 0 in i else i for i in matrix]
661+ col = [[0 ] * len (j) if 0 in j else list (j) for j in zip (* matrix)]
662+ col2row = list (map (list , zip (* col)))
663+ # 上面一行效果等同:
664+ # col2row = [list(i) for i in zip(*col)]
665+ for i in range (len (matrix)):
666+ matrix[i] = col2row[i] if row[i] != [0 ] * len (matrix[0 ]) else row[i]
667+ ```
668+ - 获取每一行 / 列的值,假如有0 就整行 / 整列置为0
669+ - 重新将列排序列表转换为行排序列表,即原来的` matrix ` 中有0的列全为0,行不变
670+ - ` zip(*col) ` 返回的是` zip ` 类型,需要转换成list,其中元素类型为元组
671+ - 所以之后做了两步转换,先将zip()返回的各个元组转换为list,在将整个转换为list
672+ - 替换matrix各行, 如果一整行为0, 则替换为0,否则为` col2row ` 对应的各行
673+ 674+ ## [ 74. 搜索二维矩阵 4行] ( https://leetcode-cn.com/problems/search-a-2d-matrix/ )
675+ ``` python
676+ class Solution :
677+ def searchMatrix (self matrix : List[List[int ]], target : int ) -> bool :
678+ if not matrix or not matrix[0 ]: return False
679+ col = list (list (zip (* matrix))[0 ]) # set() -> list()
680+ index = bisect.bisect_left(col, target, 0 , len (matrix)- 1 ) # 二分查找
681+ return target in (matrix[index] + matrix[index- 1 ])
682+ ```
683+ - 先获取首列,然后二分类找到这个数所在的行,然后进行判断
684+ 685+ 655686## [ 78. Subsets 2行] ( https://leetcode.com/problems/subsets/ )
656687``` python
657688class Solution :
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