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[LeetCode] 304. Range Sum Query 2D - Immutable

chinafgj 發布于2019-08-16 13:43 / 2108人閱讀

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).

https://leetcode.com/static/i...

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

class NumMatrix {
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return;
        }
        int m = matrix.length, n = matrix[0].length;
        this.sum = new int[m+1][n+1];

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                sum[i][j] = sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i-1][j-1];
            }
        }
    }
    
    public 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];
    }
}

GPU云服務器云服務器 Immutable 返回304 LeetCode js304

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