leetcode 304. Range Sum Query 2D - Immutable

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

 

 

此题是303的进化，303是一维数组，本题为二维数组。思路一样，只是计算时要是几个矩形的计算。

public class NumMatrix {
     private int[][] dp;


    public NumMatrix(int[][] matrix) {
          if (matrix == null || matrix.length == 0) {
            return;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        dp = new int[m + 1][n + 1];
        for (int i = 0; i < m; i++) {
            dp[i][0] = 0;
        }
        for (int i = 0; i < n; i++) {
            dp[0][1] = 0;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = matrix[i - 1][j-1 ] + dp[i ][j-1] + dp[i-1][j] - dp[i - 1][j - 1];
            }
        }
    }
    
        public int sumRegion(int row1, int col1, int row2, int col2) {
           if (dp == null) {
            return 0;
        }
        return dp[row2 + 1][col2 + 1] - dp[row2 + 1][col1] - dp[row1][col2 + 1] + dp[row1][col1];
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */