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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.com/problems/range-sum-query-2d-immutable/

解法

动态规划-二维前缀和。

Python3

class NumMatrix:

    def __init__(self, matrix: List[List[int]]):
        m, n = len(matrix), len(matrix[0])
        self.pre = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                self.pre[i][j] = self.pre[i - 1][j] + self.pre[i][j - 1] - self.pre[i - 1][j - 1] + matrix[i - 1][j - 1]

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


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

Java

class NumMatrix {
    private int[][] pre;

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

    public int sumRegion(int row1, int col1, int row2, int col2) {
        return pre[row2 + 1][col2 + 1] - pre[row2 + 1][col1] - pre[row1][col2 + 1] + pre[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);
 */

C++

class NumMatrix {
public:
    vector<vector<int>> pre;

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

    int sumRegion(int row1, int col1, int row2, int col2) {
        return pre[row2 + 1][col2 + 1] - pre[row2 + 1][col1] - pre[row1][col2 + 1] + pre[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);
 */

Go

type NumMatrix struct {
	pre [][]int
}

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

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

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

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