Day11-Flood Fill(easy)
问题描述:
An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).
Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, “flood fill” the image.
To perform a “flood fill”, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.
At the end, return the modified image.
填色游戏,给一个起点和一个新值,让我们找到这个起始点周围(上下左右)与他颜色相同的涂上新颜色,然后找到的周围值再递归的影响到其他位置。
Example:
Example 1:
Input:
image = [[1,1,1],[1,1,0],[1,0,1]]
sr = 1, sc = 1, newColor = 2
Output: [[2,2,2],[2,2,0],[2,0,1]]
Explanation:
From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected
by a path of the same color as the starting pixel are colored with the new color.
Note the bottom corner is not colored 2, because it is not 4-directionally connected
to the starting pixel.
解法:
这道题很明显的做法就是用递归了,也可以说是DFS的算法,如果当前位置为指定的颜色,那我们就染色然后在遍历其上下左右的位置,下面是我写的代码不知道哪错了,就是通过不了,说是那个helper()函数的定义位置处有个语法错误,搞不明白。。。是英文符号。。
class Solution:
def floodFill(self, image: List[List[int]], sr: int, sc: int, newColor: int) -> List[List[int]]:
def helper(image,new_color,old_color,(row,col)):
if row >= 0 and row <= len(image) and col >= 0 and col <= len(image[0]):
if image[row][col] == old_color:
image[row][col] = new_color
helper(image,new_color,old_color,(row + 1,col))
helper(image,new_color,old_color,(row,col + 1))
helper(image,new_color,old_color,(row,col - 1))
helper(image,new_color,old_color,(row - 1,col))
return image
return
return
return helper(image,new_Color,image[sr,sc],(sr,sc))
本文详细解析了LeetCode上的FloodFill(洪水填充)算法题目,介绍了如何使用递归(深度优先搜索)来解决该问题,并给出了Python实现的示例代码。通过本教程,读者可以了解FloodFill算法的基本原理及其实现细节。
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