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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(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.

维护一个和数组sums[i][j]表示以(i,j)为右下角，(0,0)为左上角的矩形中所有数字的和。sum[i][j]可以通过DP来求。

sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]  1<=i<m , i<=j<n

求范围的和，借鉴的思想是平面直角坐标系中给定矩形的右上角、左下角，利用坐标系的关系求面积的思想。

long[][] sums;boolean bad=false;public NumMatrix(int[][] matrix){int m=matrix.length;if(m<1){bad=true;return ;}int n=matrix[0].length;if(n<1){bad=true;return ;}sums=new long[m][n];for(int i=0;i<m;i++)for(int j=0;j<n;j++){long sum=matrix[i][j];if(j>=1)sum+=sums[i][j-1];if(i>=1)sum+=sums[i-1][j];if(i>=1&&j>=1)sum-=sums[i-1][j-1];sums[i][j]=sum;}}public int sumRegion(int row1, int col1, int row2, int col2){if(bad)return 0;return (int)(sums[row2][col2]-(col1>=1?sums[row2][col1-1]:0)-(row1>=1?sums[row1-1][col2]:0)+((row1>=1&&col1>=1)?sums[row1-1][col1-1]:0));}

update 2016.07.29

边界如果要单独考虑的时候，不妨考虑平移一下再dp

class NumMatrix{long[][] dp;public NumMatrix(int[][] matrix){if(matrix.length<1||matrix[0].length<1)return ;int m=matrix.length;int n=matrix[0].length;dp=new long[m+1][n+1];for(int i=1;i<=m;i++)for(int j=1;j<=n;j++)dp[i][j]=matrix[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];}public int sumRegion(int row1, int col1, int row2, int col2){return (int)(dp[row2+1][col2+1]-dp[row2+1][col1]-dp[row1][col2+1]+dp[row1][col1]);}}