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社区首页 >专栏 >LeetCode 0304 - Range Sum Query 2D - Immutable

LeetCode 0304 - Range Sum Query 2D - Immutable

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Reck Zhang
发布2021-08-11 11:53:01
发布2021-08-11 11:53:01
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文章被收录于专栏:Reck ZhangReck Zhang

Range Sum Query 2D - Immutable

Desicription

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).

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

Example:

代码语言:javascript
复制
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

Note:

  1. You may assume that the matrix does not change.
  2. There are many calls to sumRegion function.
  3. You may assume that row1 ≤ row2 and col1 ≤ col2.

Solution

代码语言:javascript
复制
class NumMatrix {
    vector<vector<int>> dp;
public:
    explicit NumMatrix(const vector<vector<int>> &matrix) {
        if(matrix.empty()) {
            return ;
        }
        dp = vector<vector<int>>(matrix.size(), vector<int>(matrix[0].size()));
        unsigned long row = matrix.size();
        unsigned long col = matrix[0].size();
        dp[0][0] = matrix[0][0];
        for(int i = 1; i < row; i++) {
            dp[i][0] = dp[i - 1][0] + matrix[i][0];
        }
        for(int i = 1; i < col; i++) {
            dp[0][i] = dp[0][i - 1] + matrix[0][i];
        }
        for(int i = 1; i < row; i++) {
            for(int j = 1; j < col; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + matrix[i][j];
            }
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        if(dp.empty()) {
            return 0;
        }
        if(row1 == 0 && col1 == 0) {
            return dp[row2][col2];
        } else if(row1 == 0) {
            return dp[row2][col2] - dp[row2][col1 - 1];
        } else if(col1 == 0) {
            return dp[row2][col2] - dp[row1 - 1][col2];
        } else {
            return dp[row2][col2] - dp[row2][col1 - 1] - dp[row1 - 1][col2] + dp[row1 - 1][col1 - 1];
        }
    }
};
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目录
  • Range Sum Query 2D - Immutable
    • Desicription
    • Solution
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