Range Sum Query 2D - Immutable

Given a 2D matrix matrix[][], find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
Example:
Given matrix = [
  [3, 0, 1, 4, 2],
  [5, 6, 3, 2, 1],
  [1, 2, 0, 1, 5],
  [4, 1, 0, 1, 7],
  [1, 0, 3, 0, 5]
]

sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

这道题目属于二维的矩阵运算，题目中假设矩阵没有变化，对比前面的一道题 Range Sum Query - Mutable来看，我们也可以用树状数组来做，这里我们就需要构建一个二维的树状数组。具体如何构建一个树状数组请看连接中的那道题。这里先给出一个AC的代码：

public class NumMatrix {
    int[][] matrix;
    int[][] BIT;
    public NumMatrix(int[][] matrix) {
        this.matrix = matrix;
        if(matrix == null || matrix.length == 0 )   return;
        BIT = new int[matrix.length + 1][matrix[0].length + 1];
        for(int i = 0; i < matrix.length; i++)
            for(int j = 0; j < matrix[0].length; j++) 
                init(i, j, matrix[i][j]);
    }
    public void init(int i, int j, int val) {
        j ++;
        while(j <= matrix[0].length) {
            BIT[i][j] += val;
            j += (j & -j);
        }
    }
    /* implement mutable 
    public void update(int i, int j, int val) {
        int differ = val - matrix[i][j];
        matrix[i][j] = val;
        init(i, j, differ);
    } 
    */
    public int sumRegion(int row1, int col1, int row2, int col2) {
        int result = 0;
        for(int i = row1; i <= row2; i++) {
            result += getSumRange(i, col1, col2);
        }
        return result;
    }
    public int getSumRange(int i, int start, int end) {
        return getSum(i, end) - getSum(i, start - 1);
    }
    public int getSum(int i, int point) {
        point ++;
        int sum = 0;
        while(point > 0) {
            sum += BIT[i][point];
            point -= (point & -point);
        }
        return sum;
    }
}


// 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);

代码中注释的部分是update方法，对应mutable的情况。

因为树状数组的优势在于解决对变化的数据求和，在不变的情况下，优势就没有那么明显，因为构建一个树状数组也需要消耗时间和空间。这道题我们采用动态规划的思想解决比较合适。
代码如下：

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

    public int sumRegion(int row1, int col1, int row2, int col2) {
        return dp[row2 + 1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1] - dp[row1][col2 + 1];
    }
}


// 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);