lc2536.子矩阵元素加1

暴力解法：直接按照题目所示在矩阵的相应位置加一

时间复杂度：O(n2 * queries.length)

空间复杂度：O(1)

二维差分：创建二维差分数组，通过对差分数组的修改来影响原来的数组，最后还原

时间复杂度：O(n2 + queries.length)

空间复杂度：O(n2)

示例2此种情况会发生角标越界的情况，因此差分数组需要多初始化2行2列

代码

import org.junit.Test;

public class SubmatrixPlus {

    @Test
    public void test() {
        int[][] queries = new int[][]{{1, 1, 2, 2}, {0, 0, 1, 1}};
        for (int[] query : submatrixPlus(queries, 3)) {
            for (int n : query) {
                System.out.print(n + " ");
            }
            System.out.println();
        }

        int[][] queries1 = new int[][]{{0, 0, 1, 1}};
        for (int[] query : submatrixPlus(queries1, 2)) {
            for (int n : query) {
                System.out.print(n + " ");
            }
            System.out.println();
        }
    }

    //int[][] querries = {{左上角行，左上角列，右下角行，右下角列}，{左上角行，左上角列，右下角行，右下角列}}
    public static int[][] submatrixPlus(int[][] queries, int n) {
        // 构建差分数组,多初始化2行2列避免数组越界
        int[][] arr = new int[n][n];
        for (int i = 0; i < queries.length; i++) {
            arr[queries[i][0] + 1][queries[i][1] + 1]++;//第几行不等于数组的索引
            arr[queries[i][2] + 2][queries[i][1] + 1]--;
            arr[queries[i][0] + 1][queries[i][3] + 2]--;
            arr[queries[i][2] + 2][queries[i][3] + 2]++;
        }
        //还原数组
        int[][] res = new int[n + 2][n + 2];
        for (int i = 0; i < res.length; i++) {
            for (int j = 0; j < res[i].length; j++) {
                arr[i + 1][j + 1] = arr[i + 1][j + 1] + arr[i + 1][j] + arr[i][j + 1] - arr[i][j];
                res[i][j] = arr[i + 1][j + 1];
            }
        }
        return res;

    }
}