leetcode 304. Range Sum Query 2D
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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).
Range Sum Query 2D
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.
题意很简单,做法也很简单,把上一道题leetcode 303. Range Sum Query - Immutable 字串求和 + DP 做一下扩展即可。
代码如下:
/* * 就是面积的迭代计算 * DP的一个简单应用 * */class NumMatrix { public int[][] sum=null; public NumMatrix(int[][] matrix) { if(matrix==null || matrix.length<=0) return ; sum=new int[matrix.length][matrix[0].length]; sum[0][0]=matrix[0][0]; for(int i=1;i<matrix.length;i++) sum[i][0]=sum[i-1][0]+matrix[i][0]; for(int i=1;i<matrix[0].length;i++) sum[0][i]=sum[0][i-1]+matrix[0][i]; for(int i=1;i<matrix.length;i++) { for(int j=1;j<matrix[0].length;j++) { sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i][j]; } } } public int sumRegion(int row1, int col1, int row2, int col2) { if(sum==null || sum.length<=0 || row1<0 || col1>=sum[0].length || row2<0 || col2>=sum[0].length) return 0; if(row1>row2 || col1 > col2) return 0; else if(row1==0 && col1==0) return sum[row2][col2]; else if(row1==0 && col1!=0) return sum[row2][col2]-sum[row2][col1-1]; else if(row1!=0 && col1==0) return sum[row2][col2]-sum[row1-1][col2]; else return sum[row2][col2]-sum[row2][col1-1]-sum[row1-1][col2]+sum[row1-1][col1-1]; }}/** * Your NumMatrix object will be instantiated and called as such: * NumMatrix obj = new NumMatrix(matrix); * int param_1 = obj.sumRegion(row1,col1,row2,col2); */
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