[LeetCode] 308. Range Sum Query 2D - Mutable 二維區域和檢索 - 可變

 

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).

The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

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
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

Note:

1. The matrix is only modifiable by the update function.
2. You may assume the number of calls to update and sumRegion function is distributed evenly.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

 

這道題讓我們求二維區域和檢索，而且告訴我們數組中的值可能變化，這是之前那道Range Sum Query 2D - Immutable的拓展，由於我們之前做過一維數組的可變和不可變的情況Range Sum Query - Mutable和Range Sum Query - Immutable，那么為了能夠通過OJ，我們還是需要用到樹狀數組Binary Indexed Tree(參見Range Sum Query - Mutable)，其查詢和修改的復雜度均為O(logn)，那么我們還是要建立樹狀數組，我們根據數組中的每一個位置，建立一個二維的樹狀數組，然后還需要一個getSum函數，以便求得從(0, 0)到(i, j)的區間的數字和，然后在求某一個區間和時，就利用其四個頂點的區間和關系可以快速求出，參見代碼如下：

 

解法一：

// Binary Indexed Tree 
class NumMatrix {
public:
    NumMatrix(vector<vector<int>> &matrix) {
        if (matrix.empty() || matrix[0].empty()) return;
        mat.resize(matrix.size() + 1, vector<int>(matrix[0].size() + 1, 0));
        bit.resize(matrix.size() + 1, vector<int>(matrix[0].size() + 1, 0));
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = 0; j < matrix[i].size(); ++j) {
                update(i, j, matrix[i][j]);
            }
        }
    }

    void update(int row, int col, int val) {
        int diff = val - mat[row + 1][col + 1];
        for (int i = row + 1; i < mat.size(); i += i&-i) {
            for (int j = col + 1; j < mat[i].size(); j += j&-j) {
                bit[i][j] += diff;
            }
        }
        mat[row + 1][col + 1] = val;
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        return getSum(row2 + 1, col2 + 1) - getSum(row1, col2 + 1) - getSum(row2 + 1, col1) + getSum(row1, col1);
    }
    
    int getSum(int row, int col) {
        int res = 0;
        for (int i = row; i > 0; i -= i&-i) {
            for (int j = col; j > 0; j -= j&-j) {
                res += bit[i][j];
            }
        }
        return res;
    } 
    
private:
    vector<vector<int>> mat;
    vector<vector<int>> bit;
};

 

我在網上還看到了另一種解法，這種解法並沒有用到樹狀數組，而是利用了列之和，所謂列之和，就是(i, j)就是(0, j) + (1, j) + ... + (i, j) 之和，相當於把很多個一維的區間之和拼到了一起，那么我們在構造函數中需要建立起這樣一個列之和矩陣，然后再更新某一個位置時，我們只需要將該列中改變的位置下面的所有數字更新一下即可，而在求某個區間和時，只要將相差的各列中對應的起始和結束的行上的值的差值累加起來即可，參見代碼如下：

 

解法二：

// Column Sum
class NumMatrix {
public:
    NumMatrix(vector<vector<int>> &matrix) {
        if (matrix.empty() || matrix[0].empty()) return;
        mat = matrix;
        colSum.resize(matrix.size() + 1, vector<int>(matrix[0].size(), 0));
        for (int i = 1; i < colSum.size(); ++i) {
            for (int j = 0; j < colSum[0].size(); ++j) {
                colSum[i][j] = colSum[i - 1][j] + matrix[i - 1][j];
            }
        }
    }

    void update(int row, int col, int val) {
        for (int i = row + 1; i < colSum.size(); ++i) {
            colSum[i][col] += val - mat[row][col];
        }
        mat[row][col] = val;
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        int res = 0;
        for (int j = col1; j <= col2; ++j) {
            res += colSum[row2 + 1][j] - colSum[row1][j];
        } 
        return res;
    }

private:
    vector<vector<int>> mat;
    vector<vector<int>> colSum;
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/308

 

類似題目：

Range Sum Query 2D - Immutable

Range Sum Query - Mutable

Range Sum Query - Immutable

 

參考資料：

https://leetcode.com/problems/range-sum-query-2d-mutable/

https://leetcode.com/problems/range-sum-query-2d-mutable/discuss/75852/15ms-easy-to-understand-java-solution

https://leetcode.com/problems/range-sum-query-2d-mutable/discuss/75870/Java-2D-Binary-Indexed-Tree-Solution-clean-and-short-17ms

 

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