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).

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];
    }
}

标签：row2,matrix,int,col2,sum,304,Immutable,sumRegion,Sum
来源： https://www.cnblogs.com/xiaobaituyun/p/10887010.html