zoukankan      html  css  js  c++  java
  • 【leetcode】1030. Matrix Cells in Distance Order

    题目如下:

    We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

    Additionally, we are given a cell in that matrix with coordinates (r0, c0).

    Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

    Example 1:

    Input: R = 1, C = 2, r0 = 0, c0 = 0
    Output: [[0,0],[0,1]]
    Explanation: The distances from (r0, c0) to other cells are: [0,1]
    

    Example 2:

    Input: R = 2, C = 2, r0 = 0, c0 = 1
    Output: [[0,1],[0,0],[1,1],[1,0]]
    Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
    The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.
    

    Example 3:

    Input: R = 2, C = 3, r0 = 1, c0 = 2
    Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
    Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
    There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].
    

    Note:

    1. 1 <= R <= 100
    2. 1 <= C <= 100
    3. 0 <= r0 < R
    4. 0 <= c0 < C

    解题思路:BFS,就这样。

    代码如下:

    class Solution(object):
        def allCellsDistOrder(self, R, C, r0, c0):
            """
            :type R: int
            :type C: int
            :type r0: int
            :type c0: int
            :rtype: List[List[int]]
            """
            visit = []
            for i in range(R):
                visit.append([0] * C)
            direction = [(1,0),(-1,0),(0,1),(0,-1)]
            queue = [(r0,c0)]
            res = []
            while len(queue) > 0:
                x,y = queue.pop(0)
                if visit[x][y] == 1:
                    continue
                res.append([x, y])
                visit[x][y] = 1
                for (i,j) in direction:
                    if x + i >= 0 and x + i < R and y + j >=0 and y+j < C and visit[x+i][y+j] == 0:
                        queue.append((x+i,y+j))
            return res
  • 相关阅读:
    STC项目风险分析
    “四则运算练习器”的开发心得与优化方案
    针对“订餐系统”的分析、改进建议与阅读心得
    记一次leetcode翻车之路---给自己做个记录
    数据库基础(一)MYSQL
    面试题第1发---记20年某次面试
    go之“hello word”
    利用python简单实现unittest
    web自动化(python)——selenium工具基本使用
    2015最强java开源oa源码
  • 原文地址:https://www.cnblogs.com/seyjs/p/10765540.html
Copyright © 2011-2022 走看看