733. Flood Fill（python+cpp）

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 of image and image[0] will be in the range [1, 50].
The given starting pixel will satisfy0 <= sr < image.length and 0 <= sc <image[0].length.
The value of each color inimage[i][j] and newColor will be an integer in[0, 65535].

解释：
dfs，思想很简单，对原始的image实现一个dfs，继续进行dfs的条件是：
1.下一个元素在范围内
2.下一个元素与当前元素的颜色相同
有一个trick就是：当新的颜色和当前正在处理的位置的颜色相同时，这时候将会有一个无限循环（因为此时image[row][col]==orig_color永远成立）。所以如果新的颜色和当前正在处理的位置的颜色相同，我们什么都不需要做，只需要返回image即可。
深度优先一个很重要的问题就是访问过的数据需要被有效标记，避免二次访问，当newcolor跟原始数据一致的时候，改变原始数据并不能产生想要的标记效果，导致无法跳出递归的循环。(当新颜色和旧颜色不一样时，改变了颜色之后会跳出循环（主要是image[row][col]!=orig_color的作用），但是一样时，就不会跳出循环了（不会跳出循环的意思是，在dfs点(i,j)周围的时候，又会遍历到点(i,j)，整个程序就一直循环啊循环啊，永不停息～～））。
python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]]
        """
        if not image:
            return image
        rows,cols,orig_color=len(image),len(image[0]),image[sr][sc]
        def dfs(row,col):
            if (not (0<=row<rows and 0<=col<cols) or image[row][col]!=orig_color):
                return
            image[row][col]=newColor
            [dfs(row+x,col+y) for x,y in ((0,1),(1,0),(0,-1),(-1,0)) ]
        if orig_color!=newColor:
            dfs(sr,sc)
        return image

c++代码：

class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
        int ori_color=image[sr][sc];
        if (ori_color!=newColor)
            dfs(sr,sc,image,ori_color,newColor);
        return image;
    
    }
    void dfs(int r,int c,vector<vector<int>>&images,int ori_color,int newColor)
    {
        vector<vector<int>> directions={{0,1},{0,-1},{1,0},{-1,0}};
        int rows=images.size();
        int cols=images[0].size();
        if (r>=rows ||r<0 ||c>=cols ||c<0 or images[r][c]!=ori_color)
            return;
        images[r][c]=newColor;
        for(auto direction:directions)
            dfs(r+direction[0],c+direction[1],images,ori_color,newColor);
    }
};

总结：
对于dfs的问题，已经遍历过的位置一定要合理标记，这样就能防止一直搜索已经搜索过的地方了。

733. Flood Fill（python+cpp）

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