304. Range Sum Query 2D - Immutable

最新推荐文章于 2024-08-08 07:21:37 发布

原创最新推荐文章于 2024-08-08 07:21:37 发布 · 343 阅读

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Problem

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:

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:

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

Solution

和 Range Sum Query - Immutable 很像。

就是用一个二维数组，其中每个元素 dp[i][j] 表示以matrix[0][0]为左上角，matrix[i][j]以右下角组成的矩形中所有元素的和，

再用 arrSum[j] 表示该行从 0 到 j 所有元素的和，

这样 dp[i][j] = dp[i-1][j] + arrSum[j]

class NumMatrix {
    vector<vector<int>> sumMatrix;
public:
    NumMatrix(vector<vector<int>> &matrix) {
        if(matrix.empty() || matrix[0].empty()) return;
        
        const int M = matrix.size(), N = matrix[0].size();
        sumMatrix.resize(M, vector<int>(N, 0));
        
        sumMatrix[0][0] = matrix[0][0];
        for( int j = 1; j < N; j++){
            sumMatrix[0][j] = sumMatrix[0][j-1] + matrix[0][j];
        }
        
        for( int i = 1; i < M ; i++){
            vector<int> sumArr(N, 0);
            sumArr[0] = matrix[i][0];
            for( int j = 0; j < N ; j++){
                if(j != 0) sumArr[j] = sumArr[j - 1] + matrix[i][j];
                sumMatrix[i][j] = sumMatrix[i-1][j] + sumArr[j];
            }
        }
    }
    
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        if(col1 == 0 && row1 == 0) return sumMatrix[row2][col2];
        if(col1 == 0) return sumMatrix[row2][col2] - sumMatrix[row1-1][col2];
        if(row1 == 0) return sumMatrix[row2][col2] - sumMatrix[row2][col1-1];
        
        return sumMatrix[row2][col2] - sumMatrix[row2][col1 -1] - sumMatrix[row1-1][col2] + sumMatrix[row1-1][col1-1];
    }
};

看了这个答案之后 https://leetcode.com/course/chapters/leetcode-101/range-sum-query-2d-immutable/

发现我的 code 还可以简化很多，

dp的大小用 M + 1 行和 N ＋ 1列来表示，其中左左边的一列和最上面的一行用零来表示 boarder,

这样 dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + matrix[i-1][j-1] ，i 和 j分别都从1开始累加久好啦～～～～

这样在算 sumRegion时，也没有那么多 corner case 啦～～～～