【前缀和】B002_LC_矩阵区域和（暴力 / 二维前缀和）

一、Problem

二、Solution

方法一：暴力

一开始把 dir 数组也弄出来的了，这是错误思想，因为这里需要将 tx 与 ty 一起判断，而一起判断会导致一个不合法就 pass 掉了，其实没必要，直接用 min、max 来取出当前座标能到达的最远座标即可。

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

class Solution {
    public int[][] matrixBlockSum(int[][] mat, int K) {
    	int r = mat.length, c = mat[0].length, s[][] = new int[r][c];
    	
    	for (int i = 0; i < r; i++)
		for (int j = 0; j < c; j++) {
            int lowi = Math.max(i-K, 0), higi = Math.min(i+K, r-1);
            int lowj = Math.max(j-K, 0), higj = Math.min(j+K, c-1);
            for (int ii = lowi; ii <= higi; ii++)
            for (int jj = lowj; jj <= higj; jj++)
    			s[i][j] += mat[ii][jj];
		}
    	return s;
    }
}

复杂度分析

• 时间复杂度：$O(n^3)$，
• 空间复杂度：$O(n^3)$，

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方法二：二维前缀和

得到前缀和后，第一感觉是求出右下角座标就可以求出答案，疏忽了左下角的座标。

class Solution {
    public int[][] matrixBlockSum(int[][] g, int K) {
        int r = g.length, c = g[0].length, s[][] = new int[r+1][c+1];
        for (int i = 1; i <= r; i++)
    	for (int j = 1; j <= c; j++) {
    		s[i][j] = g[i-1][j-1] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];
    	}
    	int[][] a=  new int[r][c];
    	for (int i = 0; i < r; i++)
		for (int j = 0; j < c; j++) {
			int mx = Math.min(i + K, r-1), my = Math.min(j + K, c-1);
			a[i][j] = s[mx+1][my+1];
		}
		return a;
    }
}

class Solution {
    public int[][] matrixBlockSum(int[][] g, int K) {
        int r = g.length, c = g[0].length, s[][] = new int[r+1][c+1];
        for (int i = 1; i <= r; i++)
    	for (int j = 1; j <= c; j++) {
    		s[i][j] = g[i-1][j-1] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];
    	}
    	int[][] a = new int[r][c];
    	for (int i = 0; i < r; i++)
		for (int j = 0; j < c; j++) {
			int lx = Math.max(i - K, 0), ly = Math.max(j - K, 0);
			int rx = Math.min(i + K, r-1), ry = Math.min(j + K, c-1);
			a[i][j] = s[rx+1][ry+1] - s[rx+1][ly] - s[lx][ry+1] + s[lx][ly];
		}
		return a;
    }
}

和求子矩阵的和没啥区别，要处理左上角座标稍微麻烦一点。

复杂度分析

• 时间复杂度：$O(R × C)$，
• 空间复杂度：$O(R × C)$，