M304.二维区域和检索 - 矩阵不可变

prefix sum, https://leetcode.cn/problems/range-sum-query-2d-immutable/

二维前缀和

#include <iostream>
#include <vector>
using namespace std;

class NumMatrix
{
    vector<vector<int>> sum;

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

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

int main()
{
    cin.tie(nullptr)->sync_with_stdio(false);

    vector<vector<int>> 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}};
    NumMatrix *numMatrix = new NumMatrix(matrix);
    cout << numMatrix->sumRegion(2, 1, 4, 3) << '\n'; // return 8 (红色矩形框的元素总和)
    cout << numMatrix->sumRegion(1, 1, 2, 2) << '\n'; // return 11 (绿色矩形框的元素总和)
    cout << numMatrix->sumRegion(1, 2, 2, 4) << '\n'; // return 12 (蓝色矩形框的元素总和)
    return 0;
}

共用时10min

思路：预处理二维前缀和．时间复杂度：构造 $O(mn)$，单次查询 $O(1)$．

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

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