Leetcode - Range Sum Query 2D - Immutable

My code:

public class NumMatrix {
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        sum = new int[row][col];
        for (int i = 0; i < row; i++) {
            sum[i][0] = matrix[i][0];
            for (int j = 1; j < col; j++) {
                sum[i][j] = sum[i][j - 1] + matrix[i][j];
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        int ret = 0;
        for (int i = row1; i <= row2; i++) {
            int sum1 = (col1 == 0 ? 0 : sum[i][col1 - 1]);
            ret += sum[i][col2] - sum1;
        }
        
        return ret;
    }
}


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);

我的复杂度还是很高, 有 O(n)

看了答案，发现有 O(1) 的做法：

My code:

public class NumMatrix {
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        sum = new int[row][col];
        int total = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                total += matrix[i][j];
                sum[i][j] = total;
                if (i - 1 >= 0) {
                    sum[i][j] += sum[i - 1][j];
                }
            }
            total = 0;
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        int part1 = 0;
        if (col1 - 1 >= 0 && row1 - 1 >= 0) {
            part1 = sum[row1 - 1][col1 - 1];
        }
        
        int part2 = 0;
        if (col1 - 1 >= 0) {
            part2 = sum[row2][col1 - 1];
        }
        
        int part3 = 0;
        if (row1 - 1 >= 0) {
            part3 = sum[row1 - 1][col2];
        }
        
        return sum[row2][col2] + part1 - part2 - part3;
    }
}


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);

reference:
https://leetcode.com/articles/range-sum-query-2d-immutable/

不知道怎么就没想出来。。。
记得同学面google就碰到了这道题目。
其实想想，这道题目挺温和的，也算是原题。
如果考虑空间，那么就直接硬算，提高时间复杂度。
如果考虑时间，需要实时，那么就牺牲空间。

Anyway, Good luck, Richardo! -- 08/21/2016