LWC 60:733. Flood Fill

LWC 60:733. Flood Fill

传送门:733. Flood Fill

Problem:

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 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].

思路: BFS,从起点开始,遇到相同的像素点加入queue,从queue中取出每个像素点时,修改成最新的颜色。BFS注意已访问过的结点需要记录下来。

JAVA代码如下:

    public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
        int n = image.length;
        int m = image[0].length;

        int[][] dir = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

        Queue<int[]> queue = new ArrayDeque<>();
        queue.offer(new int[] {sr, sc});

        Set<Integer> vis = new HashSet<>();
        vis.add(sr * m + sc);
        while (!queue.isEmpty()) {
            int[] now = queue.poll();
            int color = image[now[0]][now[1]];
            image[now[0]][now[1]] = newColor;
            for (int[] d : dir) {
                int nx = now[0] + d[0];
                int ny = now[1] + d[1];
                if (nx >= 0 && nx < n && ny >= 0 && ny < m && image[nx][ny] == color && !vis.contains(nx * m + ny)) {
                    queue.offer(new int[] {nx, ny});
                    vis.add(nx * m + ny);
                }
            }
        }
        return image;
    }

Python代码如下:

    def floodFill(self, image, sr, sc, newColor):
        n = len(image)
        m = len(image[0])

        dir = [[1, 0],[-1, 0],[0, 1],[0, -1]]

        queue = []
        queue.append([sr, sc])
        vis = []
        vis.append(sr * m + sc)
        while (queue != []):
            now = queue[0]
            queue.remove(now)
            color = image[now[0]][now[1]]
            image[now[0]][now[1]] = newColor
            for i in range(len(dir)):
                nx = now[0] + dir[i][0]
                ny = now[1] + dir[i][1]
                if nx >= 0 and nx < n and ny >= 0 and ny < m and image[nx][ny] == color and nx * m + ny not in vis:
                    queue.append([nx, ny])
                    vis.append(nx * m + ny)

        return image

Python改进版:

    def floodFill(self, image, sr, sc, newColor):
        n = len(image)
        m = len(image[0])

        dir = [[1, 0],[-1, 0],[0, 1],[0, -1]]

        queue = []
        queue.append([sr, sc])
        vis = set()
        vis.add((sr, sc))
        while queue:
            now = queue.pop(0)
            color = image[now[0]][now[1]]
            image[now[0]][now[1]] = newColor
            for i in range(len(dir)):
                nx = now[0] + dir[i][0]
                ny = now[1] + dir[i][1]
                if nx >= 0 and nx < n and ny >= 0 and ny < m and image[nx][ny] == color and (nx, ny) not in vis:
                    queue.append([nx, ny])
                    vis.add((nx, ny))
        return image     

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