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].

解釋： dfs，思想很簡單，對原始的image實現一個dfs，繼續進行dfs的條件是： 1.下一個元素在範圍內 2.下一個元素與當前元素的顏色相同 有一個trick就是：當新的顏色和當前正在處理的位置的顏色相同時，這時候將會有一個無限迴圈（因為此時image[row][col]==orig_color

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++程式碼：

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);
    }
};

總結： 對於dfs的問題，已經遍歷過的位置一定要合理標記，這樣就能防止一直搜尋已經搜尋過的地方了。