34 在排序數組中查找元素的第一個和最後一個位置
Question
給定一個按照升序排列的整數數組 nums，和一個目標值 target。找出給定目標值在數組中的開始位置和結束位置。
你的算法時間複雜度必須是 O(log n) 級別。
如果數組中不存在目標值，返回 [-1, -1]。
示例 1:
輸入: nums = [5,7,7,8,8,10], target = 8
輸出: [3,4]
示例 2:
輸入: nums = [5,7,7,8,8,10], target = 6
輸出: [-1,-1]
Answer
#
# @lc app=leetcode.cn id=34 lang=python3
#
# [34] 在排序數組中查找元素的第一個和最後一個位置
#
# @lc code=start
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
start = 0
end = len(nums) - 1
# 空數組的情況
if end < 0:
return [-1, -1]
# 二分查找target落在的位置
while (start < end):
if nums[start] == target:
end = start
break
if nums[end] == target:
start = end
break
mid = (start + end) // 2
if nums[mid] < target:
start = mid + 1
if nums[mid] > target:
end = mid - 1
if nums[mid] == target:
start = mid
end = mid
break
# 如果找不到target
if nums[start] != target:
return [-1, -1]
# 找到target，向前向後搜索邊界
while (start > 0 and nums[start - 1] == target):
start -= 1
while (end < len(nums) - 1 and nums[end + 1] == target):
end += 1
return [start, end]
# @lc code=end
36 有效的數獨
Question
判斷一個 9x9 的數獨是否有效。只需要根據以下規則，驗證已經填入的數字是否有效即可。
- 數字
1-9在每一行只能出現一次。 - 數字
1-9在每一列只能出現一次。 - 數字
1-9在每一個以粗實線分隔的3x3宮內只能出現一次。
上圖是一個部分填充的有效的數獨。
數獨部分空格內已填入了數字，空白格用 '.' 表示。
示例 1:
輸入:
[
["5","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]
輸出: true
示例 2:
輸入:
[
["8","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]
輸出: false
解釋: 除了第一行的第一個數字從 5 改爲 8 以外，空格內其他數字均與 示例1 相同。
但由於位於左上角的 3x3 宮內有兩個 8 存在, 因此這個數獨是無效的。
說明:
- 一個有效的數獨（部分已被填充）不一定是可解的。
- 只需要根據以上規則，驗證已經填入的數字是否有效即可。
- 給定數獨序列只包含數字
1-9和字符'.'。 - 給定數獨永遠是
9x9形式的。
Answer
#
# @lc app=leetcode.cn id=36 lang=python3
#
# [36] 有效的數獨
#
# @lc code=start
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
import numpy as np
board = np.array(board)
row = np.zeros((10, 10))
column = np.zeros((10, 10))
group = np.zeros((10, 10))
for i in range(9):
for j in range(9):
if board[i, j] == '.':
continue
num = int(board[i, j])
if row[i][num] == 0 and column[j][num] == 0 and group[i // 3 * 3 + j // 3][num] == 0:
row[i][num] = 1
column[j][num] = 1
group[i // 3 * 3 + j // 3][num] = 1
else:
return False
return True
# @lc code=end
39 組合總和
Question
給定一個無重複元素的數組 candidates 和一個目標數 target ，找出 candidates 中所有可以使數字和爲 target 的組合。
candidates 中的數字可以無限制重複被選取。
說明：
- 所有數字（包括
target）都是正整數。 - 解集不能包含重複的組合。
示例 1:
輸入: candidates = [2,3,6,7], target = 7,
所求解集爲:
[
[7],
[2,2,3]
]
示例 2:
輸入: candidates = [2,3,5], target = 8,
所求解集爲:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
Answer
#
# @lc app=leetcode.cn id=39 lang=python3
#
# [39] 組合總和
#
# @lc code=start
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
size = len(candidates)
if size == 0:
return []
# 剪枝是爲了提速，在本題非必需
candidates.sort()
# 在遍歷的過程中記錄路徑，它是一個棧
path = []
res = []
# 注意要傳入 size ，在 range 中， size 取不到
self.__dfs(candidates, 0, size, path, res, target)
return res
def __dfs(self, candidates, begin, size, path, res, target):
# 先寫遞歸終止的情況
if target == 0:
# Python 中可變對象是引用傳遞，因此需要將當前 path 裏的值拷貝出來
# 或者使用 path.copy()
res.append(path[:])
return
for index in range(begin, size):
residue = target - candidates[index]
# “剪枝”操作，不必遞歸到下一層，並且後面的分支也不必執行
if residue < 0:
break
path.append(candidates[index])
# 因爲下一層不能比上一層還小，起始索引還從 index 開始
self.__dfs(candidates, index, size, path, res, residue)
path.pop()
# @lc code=end