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:
- 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.
解题思路:
dynamic programming m 为matrix 行, n 为matrix 列 注意matrix 为空的情况。
建立accu = new int[m+1][n+1], 第一行和第一列为0
每个坐标的值 recurrence formula: accu[i][j] = accu[i-1][j] + accu[i][j-1] - accu[i-1][j-1] + matrix[i-1][j-1]
所选长方形区间画图可知由4个顶点决定值,
公式:accu[row2+1][col2+1] - accu[row2+1][col1] - accu[row1][col2+1] + accu[row1][col1]
Java code:
public class NumMatrix { private int[][] accu; public NumMatrix(int[][] matrix) { if(matrix == null || matrix.length == 0) { return; } int m = matrix.length; int n = matrix[0].length; accu = new int[m+1][n+1]; //base case first row/column is 0 default is 0 for(int i = 1; i<= m; i++){ for(int j = 1; j <= n; j++){ accu[i][j] = accu[i-1][j] + accu[i][j-1] - accu[i-1][j-1] + matrix[i-1][j-1]; } } } public int sumRegion(int row1, int col1, int row2, int col2) { return accu[row2+1][col2+1] - accu[row2+1][col1] - accu[row1][col2+1] + accu[row1][col1]; } } // Your NumMatrix object will be instantiated and called as such: // NumMatrix numMatrix = new NumMatrix(matrix); // numMatrix.sumRegion(0, 1, 2, 3); // numMatrix.sumRegion(1, 2, 3, 4);
Reference:
1. https://leetcode.com/discuss/69054/dp-java-solution
2. https://leetcode.com/discuss/69315/an-easy-c-dp-solution-with-simple-explanation