leetcode [304]Range Sum Query 2D - Immutable

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).

leetcode [304]Range Sum Query 2D - Immutable
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:

  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.

题目大意:

求解一个矩阵范围内的数字和。

解法:

构造一个二维数组和[sum+1][col+1],sum[i+1][j+1]表示从matrix[0][0]到matrix[i][j]的数字之和。

class NumMatrix {
    private int m,n;
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        m=matrix.length;
        n=(m==0?0:matrix[0].length);
        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[row1][col2+1]-sum[row2+1][col1]+sum[row1][col1];
    }
}

  

上一篇:python教程93--python综合案例分析


下一篇:pandas 中 to_dict 的用法