Queue and BFS

發表於 Disqus：
文章字數： 4.9k 所需閱讀時間 ≈ 9 分鐘

本文蒐集

LeetCode 200: Number of Islands

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1: 

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1
Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3
Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • grid[i][j] is '0' or '1'.

我的答案

class Solution(object):
    def is_valid(self, grid, r, c):
        m, n = len(grid), len(grid[0])
        if r < 0 or c < 0 or r >= m or c >= n:
            return False
        return True
    def numIslandsBFS(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if not grid or not grid[0]:
            return 0

        m, n = len(grid), len(grid[0])
        count = 0
        for i in xrange(m):
            for j in xrange(n):
                if grid[i][j] == '1':
                    self.bfs(grid, i, j)
                    count += 1
        return count

    def bfs(self, grid, r, c):
        queue = collections.deque()
        queue.append((r, c))
        grid[r][c] = '0'
        while queue:
            directions = [(0,1), (0,-1), (-1,0), (1,0)]
            r, c = queue.popleft()
            for d in directions:
                nr, nc = r + d[0], c + d[1]    
                if self.is_valid(grid, nr, nc) and grid[nr][nc] == '1':
                    queue.append((nr, nc))
                    grid[nr][nc] = '0'

LeetCode 752: Open the Lock

You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.

The lock initially starts at '0000', a string representing the state of the 4 wheels.

You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.

Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.

Example 1: 

Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
Output: 6
Explanation:
A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
because the wheels of the lock become stuck after the display becomes the dead end "0102".
Example 2: 
Input: deadends = ["8888"], target = "0009"
Output: 1
Explanation:
We can turn the last wheel in reverse to move from "0000" -> "0009".
Example 3: 
Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
Output: -1
Explanation:
We can't reach the target without getting stuck.
Example 4: 
Input: deadends = ["0000"], target = "8888"
Output: -1

Constraints:

  • 1 <= deadends.length <= 500
  • deadends[i].length == 4
  • target.length == 4
  • target will not be in the list deadends.
  • target and deadends[i] consist of digits only.

我的解答

from collections import deque

class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
        
        if '0000' in deadends:
            return -1
        
        queue, visited = deque(['0000']), set(deadends)
        count = -1
        
        while len(queue) != 0:
            count += 1
            size = len(queue)
            
            for _ in range(size):
                
                curr = queue.popleft()

                if curr == target:
                    return count

                if curr in visited:
                    continue

                visited.add(curr)
                
                res = []
                for i, char in enumerate(curr):
                    num = int(char)
                    res.append(curr[:i] + str((num - 1) % 10) + curr[i+1:])
                    res.append(curr[:i] + str((num + 1) % 10) + curr[i+1:])
                
                queue.extend(res)
                
        return -1

LeetCode 279: Perfect Squares

Given an integer n, return the least number of perfect square numbers that sum to n.

A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.

Example 1: 

Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2: 
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
Constraints:

  • 1 <= n <= \(10^4\)

我的解答

class Solution:
    def numSquares(self, n: int) -> int:
        if n < 2:
            return n
        lst = []
        i = 1
        while i * i <= n:
            lst.append( i * i )
            i += 1
        cnt = 0
        toCheck = {n}
        while toCheck:
            cnt += 1
            temp = set()
            for x in toCheck:
                for y in lst:
                    if x == y:
                        return cnt
                    if x < y:
                        break
                    temp.add(x-y)
            toCheck = temp