图像渲染

733. 图像渲染

有一幅以 m x n 的二维整数数组表示的图画 image ，其中 image[i][j] 表示该图画的像素值大小。

你也被给予三个整数 sr , sc 和 newColor 。你应该从像素 image[sr][sc] 开始对图像进行上色填充。

为了完成 上色工作 ，从初始像素开始，记录初始坐标的 上下左右四个方向上 像素值与初始坐标相同的相连像素点，接着再记录这四个方向上符合条件的像素点与他们对应 四个方向上 像素值与初始坐标相同的相连像素点，……，重复该过程。将所有有记录的像素点的颜色值改为 newColor 。

最后返回 经过上色渲染后的图像 。

示例 1:

输入: image = [[1,1,1],[1,1,0],[1,0,1]]，sr = 1, sc = 1, newColor = 2
输出: [[2,2,2],[2,2,0],[2,0,1]]
解析: 在图像的正中间，(坐标(sr,sc)=(1,1)),在路径上所有符合条件的像素点的颜色都被更改成2。
注意，右下角的像素没有更改为2，因为它不是在上下左右四个方向上与初始点相连的像素点。

示例 2:

输入: image = [[0,0,0],[0,0,0]], sr = 0, sc = 0, newColor = 2
输出: [[2,2,2],[2,2,2]]

提示:

m == image.length
n == image[i].length
1 <= m, n <= 50
0 <= image[i][j], newColor < 2¹⁶
0 <= sr < m
0 <= sc < n

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上次编辑到这里，代码来自缓存点击恢复默认模板

class Solution { public: vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) { } };

golang 解法, 执行用时: 4 ms, 内存消耗: 3.6 MB, 提交时间: 2023-11-17 17:44:43

var (
    dx = []int{1, 0, 0, -1}
    dy = []int{0, 1, -1, 0}
)

func floodFill(image [][]int, sr int, sc int, color int) [][]int {
    currColor := image[sr][sc]
    if currColor == color {
        return image
    }
    n, m := len(image), len(image[0])
    queue := [][]int{}
    queue = append(queue, []int{sr, sc})
    image[sr][sc] = color
    for i := 0; i < len(queue); i++ {
        cell := queue[i]
        for j := 0; j < 4; j++ {
            mx, my := cell[0] + dx[j], cell[1] + dy[j]
            if mx >= 0 && mx < n && my >= 0 && my < m && image[mx][my] == currColor {
                queue = append(queue, []int{mx, my})
                image[mx][my] = color
            }
        }
    }
    return image
}

golang 解法, 执行用时: 4 ms, 内存消耗: 3.5 MB, 提交时间: 2023-11-17 17:44:30

var (
    dx = []int{1, 0, 0, -1}
    dy = []int{0, 1, -1, 0}
)

func floodFill(image [][]int, sr int, sc int, color int) [][]int {
    currColor := image[sr][sc]
    if currColor != color {
        dfs(image, sr, sc, currColor, color)
    }
    return image
}

func dfs(image [][]int, x, y, currColor, color int) {
    if image[x][y] == currColor {
        image[x][y] = color
        for i := 0; i < 4; i++ {
            mx, my := x + dx[i], y + dy[i]
            if mx >= 0 && mx < len(image) && my >= 0 && my < len(image[0]) {
                dfs(image, mx, my, currColor, color)
            }
        }
    }
}

python3 解法, 执行用时: 56 ms, 内存消耗: 16.3 MB, 提交时间: 2023-11-17 17:44:12

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]:
        currColor = image[sr][sc]
        if currColor == color:
            return image
        
        n, m = len(image), len(image[0])
        que = collections.deque([(sr, sc)])
        image[sr][sc] = color
        while que:
            x, y = que.popleft()
            for mx, my in [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]:
                if 0 <= mx < n and 0 <= my < m and image[mx][my] == currColor:
                    que.append((mx, my))
                    image[mx][my] = color
        
        return image
        

    def floodFill2(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]:
        n, m = len(image), len(image[0])
        currColor = image[sr][sc]

        def dfs(x: int, y: int):
            if image[x][y] == currColor:
                image[x][y] = color
                for mx, my in [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]:
                    if 0 <= mx < n and 0 <= my < m and image[mx][my] == currColor:
                        dfs(mx, my)

        if currColor != color:
            dfs(sr, sc)
        return image

java 解法, 执行用时: 0 ms, 内存消耗: 42.9 MB, 提交时间: 2023-11-17 17:43:39

class Solution {
    int[] dx = {1, 0, 0, -1};
    int[] dy = {0, 1, -1, 0};

    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int currColor = image[sr][sc];
        if (currColor != color) {
            dfs(image, sr, sc, currColor, color);
        }
        return image;
    }

    public void dfs(int[][] image, int x, int y, int currColor, int color) {
        if (image[x][y] == currColor) {
            image[x][y] = color;
            for (int i = 0; i < 4; i++) {
                int mx = x + dx[i], my = y + dy[i];
                if (mx >= 0 && mx < image.length && my >= 0 && my < image[0].length) {
                    dfs(image, mx, my, currColor, color);
                }
            }
        }
    }
}

class Solution2 {
    int[] dx = {1, 0, 0, -1};
    int[] dy = {0, 1, -1, 0};

    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
        int currColor = image[sr][sc];
        if (currColor == color) {
            return image;
        }
        int n = image.length, m = image[0].length;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[]{sr, sc});
        image[sr][sc] = color;
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int x = cell[0], y = cell[1];
            for (int i = 0; i < 4; i++) {
                int mx = x + dx[i], my = y + dy[i];
                if (mx >= 0 && mx < n && my >= 0 && my < m && image[mx][my] == currColor) {
                    queue.offer(new int[]{mx, my});
                    image[mx][my] = color;
                }
            }
        }
        return image;
    }
}

cpp 解法, 执行用时: 8 ms, 内存消耗: 14.3 MB, 提交时间: 2023-11-17 17:43:12

// bfs
class Solution {
public:
    const int dx[4] = {1, 0, 0, -1};
    const int dy[4] = {0, 1, -1, 0};
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        int currColor = image[sr][sc];
        if (currColor == color) {
            return image;
        }
        int n = image.size(), m = image[0].size();
        queue<pair<int, int>> que;
        que.emplace(sr, sc);
        image[sr][sc] = color;
        while (!que.empty()) {
            int x = que.front().first, y = que.front().second;
            que.pop();
            for (int i = 0; i < 4; i++) {
                int mx = x + dx[i], my = y + dy[i];
                if (mx >= 0 && mx < n && my >= 0 && my < m && image[mx][my] == currColor) {
                    que.emplace(mx, my);
                    image[mx][my] = color;
                }
            }
        }
        return image;
    }
};

// dfs
class Solution2 {
public:
    const int dx[4] = {1, 0, 0, -1};
    const int dy[4] = {0, 1, -1, 0};
    void dfs(vector<vector<int>>& image, int x, int y, int currColor, int color) {
        if (image[x][y] == currColor) {
            image[x][y] = color;
            for (int i = 0; i < 4; i++) {
                int mx = x + dx[i], my = y + dy[i];
                if (mx >= 0 && mx < image.size() && my >= 0 && my < image[0].size()) {
                    dfs(image, mx, my, currColor, color);
                }
            }
        }
    }

    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        int currColor = image[sr][sc];
        if (currColor != color) {
            dfs(image, sr, sc, currColor, color);
        }
        return image;
    }
};

golang 解法, 执行用时: 4 ms, 内存消耗: 4 MB, 提交时间: 2022-07-19 16:36:54

func floodFill(image [][]int, sr int, sc int, newColor int) [][]int {
    helper(image, sr, sc, image[sr][sc], newColor)
    return image
}

func helper(image [][]int, i, j int, oldColor, newColor int) {
    if i < 0 || i >= len(image) || j < 0 || j >= len(image[0]) || image[i][j] != oldColor || image[i][j] == newColor {
        return
    }
    image[i][j] = newColor
    helper(image, i+1, j, oldColor, newColor)
    helper(image, i-1, j, oldColor, newColor)
    helper(image, i, j+1, oldColor, newColor)
    helper(image, i, j-1, oldColor, newColor)
}

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