[LeetCode] 733. Flood Fill 洪水填充

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].

这道题给了一个用二维数组表示的图像，不同的数字代表不同的颜色，给了一个起始点坐标，还有一个新的颜色，让我们把起始点的颜色以及其相邻的同样的颜色都换成新的颜色。实际上就是一个找相同区间的题，可以用 BFS 或者 DFS 来做。先来看 BFS 的解法，使用一个队列 queue 来辅助，首先将给定点放入队列中，然后进行 while 循环，条件是 queue 不为空，然后进行类似层序遍历的方法，取出队首元素，将其赋值为新的颜色，然后遍历周围四个点，如果不越界，且周围的颜色跟起始颜色相同的话，将位置加入队列中，参见代码如下：

解法一：

class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
        int m = image.size(), n = image[0].size(), color = image[sr][sc];
        vector<vector<int>> res = image;
        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
        queue<pair<int, int>> q{{{sr, sc}}};
        while (!q.empty()) {
            int len = q.size();
            for (int i = 0; i < len; ++i) {
                auto t = q.front(); q.pop();
                res[t.first][t.second] = newColor;
                for (auto dir : dirs) {
                    int x = t.first + dir[0], y = t.second + dir[1];
                    if (x < 0 || x >= m || y < 0 || y >= n || res[x][y] != color) continue;
                    q.push({x, y});
                }
            }
        }
        return res;
    }
};

DFS 的写法相对简洁一些，首先判断如果给定位置的颜色跟新的颜色相同的话，直接返回，否则就对给定位置调用递归函数。在递归函数中，如果越界或者当前颜色跟起始颜色不同，直接返回。否则就给当前位置赋上新的颜色，然后对周围四个点继续调用递归函数，参见代码如下：

解法二：

class Solution {
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int newColor) {
        if (image[sr][sc] == newColor) return image;
        helper(image, sr, sc, image[sr][sc], newColor);
        return image;
    }
    void helper(vector<vector<int>>& image, int i, int j, int color, int newColor) {
        int m = image.size(), n = image[0].size();
        if (i < 0 || i >= m || j < 0 || j >= n || image[i][j] != color) return;
        image[i][j] = newColor;
        helper(image, i + 1, j, color, newColor);
        helper(image, i, j + 1, color, newColor);
        helper(image, i - 1, j, color, newColor);
        helper(image, i, j - 1, color, newColor);
    }
};