python--leetcode 733. Flood Fill
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.
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.
Note:
The length ofimage and
image[0] will be in the range
[1, 50]. The given starting pixel will satisfy
0 <= sr < image.length and
0 <= sc < image[0].length.
The value of each color in
image[i][j] and
newColor will be an integer in
[0, 65535].
题目意思虽然长但是比较简单，就是给你一个二维数组，其中一种数字等于一个颜色。再给你一个坐标和一个新颜色，让你把这个坐标代表的点以及周围和它颜色一样的点全染成新颜色。
我记得之前有做过类似的一题，思路就是递归去遍历所有符合条件的点，即DFS。
上代码：
class Solution(object):
def floodFill(self, image, sr, sc, newColor):
"""
:type image: List[List[int]]
:type sr: int
:type sc: int
:type newColor: int
:rtype: List[List[int]]
"""
self.oldColor=image[sr][sc]
self.aa=[[0 for i in range(len(image[0]))] for j in range(len(image))]
def search(image,i,j):
self.aa[i][j]=1
image[i][j]=newColor
if i>=1 :
if image[i-1][j]==self.oldColor and self.aa[i-1][j]==0:
search(image,i-1,j)
if j>=1 :
if image[i][j-1]==self.oldColor and self.aa[i][j-1]==0:
search(image,i,j-1)
if i<=len(image)-2 :
if image[i+1][j]==self.oldColor and self.aa[i+1][j]==0:
search(image,i+1,j)
if j<=len(image[0])-2 :
if image[i][j+1]==self.oldColor and self.aa[i][j+1]==0:
search(image,i,j+1)
search(image,sr,sc)
return image
s=Solution()
image = [[1,1,1],[1,1,0],[1,0,1]]
print(s.floodFill(image,1,1,2))
再上一段代码，这段就优雅多了：
class Solution(object):
def floodFill(self, image, sr, sc, newColor):
rows, cols, orig_color = len(image), len(image[0]), image[sr][sc]
def traverse(row, col):
if (not (0 <= row < rows and 0 <= col < cols)) or image[row][col] != orig_color:
return
image[row][col] = newColor
[traverse(row + x, col + y) for (x, y) in ((0, 1), (1, 0), (0, -1), (-1, 0))]
if orig_color != newColor:
traverse(sr, sc)
return image