@@ -454,4 +454,118 @@ class Solution:
454454 res += edge
455455
456456 return res
457+ ```
458+ 459+ ## 第 9 场双周赛
460+ 461+ [ 点击前往第 9 场双周赛] ( https://leetcode-cn.com/contest/biweekly-contest-9 )
462+ 463+ ### 5072. 最多可以买到的苹果数量
464+ 465+ [ 原题链接] ( https://leetcode-cn.com/contest/biweekly-contest-9/problems/how-many-apples-can-you-put-into-the-basket/ )
466+ 467+ #### 思路
468+ 469+ 1 . 排序
470+ 2 . 遍历排序结果,将苹果重量依次相加:
471+ - 如果相加和小于 5000,则把这个苹果放入袋子
472+ - 如果相加和大于 5000,跳出遍历循环
473+ 474+ ``` python
475+ class Solution :
476+ def maxNumberOfApples (self arr : List[int ]) -> int :
477+ arr.sort()
478+ res = 0
479+ s = 0
480+ for n in arr:
481+ s += n
482+ if s > 5000 :
483+ break
484+ else :
485+ res += 1
486+ return res
487+ ```
488+ 489+ ### 5073. 进击的骑士
490+ 491+ [ 原题链接] ( https://leetcode-cn.com/contest/biweekly-contest-9/problems/minimum-knight-moves/ )
492+ 493+ #### 思路
494+ 495+ BFS,需要进行枝剪,否则会超时。
496+ 497+ 骑士所走的 8 个方向具有对称性,所以我们只要保留 ` x > 0 && y > 0 ` 方向即可。这里借用了队列来实现 BFS。
498+ 499+ ``` python
500+ class Solution :
501+ def minKnightMoves (self x : int , y : int ) -> int :
502+ if x == 0 and y == 0 :
503+ return 0
504+ 505+ dx = [2 , 2 , - 2 , - 2 , 1 , 1 , - 1 , - 1 ]
506+ dy = [1 , - 1 , 1 , - 1 , 2 , - 2 , 2 , - 2 ]
507+ # 对称性
508+ x = - x if x < 0 else x
509+ y = - y if y < 0 else y
510+ 511+ # 初始化队列
512+ q = list ()
513+ mark = dict ()
514+ mark[0 ] = True
515+ q.append([0 , 0 ])
516+ M = 10000
517+ 518+ res = 0
519+ while len (q):
520+ q_length = len (q)
521+ res += 1
522+ # 遍历队列现有元素
523+ while q_length:
524+ px, py = q[0 ][0 ], q[0 ][1 ]
525+ # 取出队列头部元素
526+ del (q[0 ])
527+ q_length -= 1
528+ 529+ for i in range (len (dx)):
530+ pdx = px + dx[i]
531+ pdy = py + dy[i]
532+ # 对称性枝剪
533+ if pdx < 0 or pdy < 0 :
534+ continue
535+ if abs (pdx) + abs (pdy) > 300 :
536+ continue
537+ # 是否为要求的值
538+ if pdx == x and pdy == y:
539+ return res
540+ # 是否已记录
541+ tmp = pdx * M + pdy
542+ if tmp in mark:
543+ continue
544+ mark[tmp] = True
545+ q.append([pdx, pdy])
546+ ```
547+ 548+ ### 5071. 找出所有行中最小公共元素
549+ 550+ [ 原题链接] ( https://leetcode-cn.com/contest/biweekly-contest-9/problems/find-smallest-common-element-in-all-rows/ )
551+ 552+ #### 思路
553+ 554+ 1 . 用字典 ` key ` 存储每行中存在的元素,` value ` 值为元素个数
555+ 2 . 从小到大遍历字典 ` key ` 所代表的元素,若 ` value ` 值等于 ` len(mat) ` (代表每行都出线了这个元素),则返回该元素
556+ 557+ ``` python
558+ class Solution :
559+ def smallestCommonElement (self mat : List[List[int ]]) -> int :
560+ mat_num = len (mat)
561+ mat_map = dict ()
562+ for nums in mat:
563+ for num in nums:
564+ mat_map[num] = mat_map.get(num, 0 ) + 1
565+ 566+ key_list = sorted ([x for x in mat_map])
567+ for k in key_list:
568+ if mat_map.get(k, 0 ) == mat_num:
569+ return k
570+ return - 1
457571```
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