LeetCode 中等题解（3）

34 在排序数组中查找元素的第一个和最后一个位置

Question

给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]

Answer

#
# @lc app=leetcode.cn id=34 lang=python3
#
# [34] 在排序数组中查找元素的第一个和最后一个位置
#


# @lc code=start
class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        start = 0
        end = len(nums) - 1
        # 空数组的情况
        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，向前向后搜索边界
        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 有效的数独

Question

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入:
[
  ["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"]
]
输出: true

示例 2:

输入:
[
  ["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"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

说明:

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 给定数独序列只包含数字 1-9 和字符 '.' 。
• 给定数独永远是 9x9 形式的。

Answer

#
# @lc app=leetcode.cn id=36 lang=python3
#
# [36] 有效的数独
#


# @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 组合总和

Question

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

• 所有数字（包括 target）都是正整数。
• 解集不能包含重复的组合。

示例 1:

输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
  [7],
  [2,2,3]
]

示例 2:

输入: candidates = [2,3,5], target = 8,
所求解集为:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

Answer

#
# @lc app=leetcode.cn id=39 lang=python3
#
# [39] 组合总和
#

# @lc code=start
class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        size = len(candidates)
        if size == 0:
            return []

        # 剪枝是为了提速，在本题非必需
        candidates.sort()
        # 在遍历的过程中记录路径，它是一个栈
        path = []
        res = []
        # 注意要传入 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):
        # 先写递归终止的情况
        if target == 0:
            # Python 中可变对象是引用传递，因此需要将当前 path 里的值拷贝出来
            # 或者使用 path.copy()
            res.append(path[:])
            return

        for index in range(begin, size):
            residue = target - candidates[index]
            # “剪枝”操作，不必递归到下一层，并且后面的分支也不必执行
            if residue < 0:
                break
            path.append(candidates[index])
            # 因为下一层不能比上一层还小，起始索引还从 index 开始
            self.__dfs(candidates, index, size, path, res, residue)
            path.pop()


# @lc code=end

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