剑指 Offer II 013. 二维子矩阵的和

题目描述

给定一个二维矩阵 matrix，以下类型的多个请求：

计算其子矩形范围内元素的总和，该子矩阵的左上角为 (row1, col1) ，右下角为 (row2, col2) 。

实现 NumMatrix 类：

NumMatrix(int[][] matrix) 给定整数矩阵 matrix 进行初始化
int sumRegion(int row1, int col1, int row2, int col2) 返回左上角 (row1, col1) 、右下角 (row2, col2) 的子矩阵的元素总和。

示例 1：

输入:
["NumMatrix","sumRegion","sumRegion","sumRegion"]
[[[[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]]],[2,1,4,3],[1,1,2,2],[1,2,2,4]]
输出:
[null, 8, 11, 12]

解释:
NumMatrix numMatrix = new NumMatrix([[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]]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)

提示：

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-10⁵ <= matrix[i][j] <= 10⁵
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
最多调用 10⁴ 次 sumRegion 方法

注意：本题与主站 304 题相同： https://leetcode.cn/problems/range-sum-query-2d-immutable/

解法

方法一：二维前缀和

我们可以用一个二维数组 $s$ 来保存矩阵 $matrix$ 的前缀和，其中 $s[i+1][j+1]$ 表示矩阵 $matrix$ 中以 $(0,0)$ 为左上角，$(i,j)$ 为右下角的子矩阵中所有元素的和。

那么：

$$ \begin{aligned} s[i+1][j+1] &= s[i][j+1] + s[i+1][j] - s[i][j] + matrix[i][j] \end{aligned} $$

我们可以用前缀和数组 $s$ 来快速计算矩阵 $matrix$ 中任意子矩阵的元素和，计算公式如下：

$$ \begin{aligned} &\textit{sumRegion}(row_1,col_1,row_2,col_2) \ &= s[row_2+1][col_2+1] - s[row_2+1][col_1] - s[row_1][col_2+1] + s[row_1][col_1] \end{aligned} $$

时间复杂度：

初始化的时间复杂度为 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵 $matrix$ 的行数和列数。
每次计算子矩阵的元素和的时间复杂度为 $O(1)$。

空间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵 $matrix$ 的行数和列数。我们需要创建一个二维数组 $s$ 来保存矩阵 $matrix$ 的前缀和。

Python3JavaC++GoTypeScript

class NumMatrix:
    def __init__(self, matrix: List[List[int]]):
        self.s = [[0] * (len(matrix[0]) + 1) for _ in range(len(matrix) + 1)]
        for i, row in enumerate(matrix, 1):
            for j, x in enumerate(row, 1):
                self.s[i][j] = (
                    self.s[i - 1][j] + self.s[i][j - 1] - self.s[i - 1][j - 1] + x
                )

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        return (
            self.s[row2 + 1][col2 + 1]
            - self.s[row2 + 1][col1]
            - self.s[row1][col2 + 1]
            + self.s[row1][col1]
        )


# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# param_1 = obj.sumRegion(row1,col1,row2,col2)

class NumMatrix {
    private int[][] s;

    public NumMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        s = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + matrix[i - 1][j - 1];
            }
        }
    }

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

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

class NumMatrix {
public:
    NumMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        s.resize(m + 1, vector<int>(n + 1, 0));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + matrix[i - 1][j - 1];
            }
        }
    }

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

private:
    vector<vector<int>> s;
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix* obj = new NumMatrix(matrix);
 * int param_1 = obj->sumRegion(row1,col1,row2,col2);
 */

type NumMatrix struct {
    s [][]int
}

func Constructor(matrix [][]int) NumMatrix {
    m, n := len(matrix), len(matrix[0])
    s := make([][]int, m+1)
    for i := 0; i < m+1; i++ {
        s[i] = make([]int, n+1)
    }
    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            s[i][j] = s[i-1][j] + s[i][j-1] + -s[i-1][j-1] + matrix[i-1][j-1]
        }
    }
    return NumMatrix{s}
}

func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int {
    return this.s[row2+1][col2+1] - this.s[row2+1][col1] - this.s[row1][col2+1] + this.s[row1][col1]
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * obj := Constructor(matrix);
 * param_1 := obj.SumRegion(row1,col1,row2,col2);
 */

class NumMatrix {
    s: number[][];

    constructor(matrix: number[][]) {
        const m = matrix.length;
        const n = matrix[0].length;
        this.s = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
        for (let i = 1; i <= m; i++) {
            for (let j = 1; j <= n; j++) {
                this.s[i][j] =
                    this.s[i - 1][j] +
                    this.s[i][j - 1] -
                    this.s[i - 1][j - 1] +
                    matrix[i - 1][j - 1];
            }
        }
    }

    sumRegion(row1: number, col1: number, row2: number, col2: number): number {
        return (
            this.s[row2 + 1][col2 + 1] -
            this.s[row2 + 1][col1] -
            this.s[row1][col2 + 1] +
            this.s[row1][col1]
        );
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * var obj = new NumMatrix(matrix)
 * var param_1 = obj.sumRegion(row1,col1,row2,col2)
 */

剑指 Offer II 013. 二维子矩阵的和

题目描述

解法

方法一：二维前缀和

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