303+304. Range Sum Query && Range Sum Query 2D

题目：

Given an integer array nums,
find the sum of the elements between indices i and j (i ≤ j),
inclusive.

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).

思路：

一个非常简单的dp，即dp[i]表示数组从第1个到第i个的和。

还是一个非常简单的dp，即dp[i][j]表示以（i,j）为右下标的长方体内所有数字的和，然而这道题因为null的情况好几次都没有通过！

程序：

public class NumArray {
private int[] dp;

public NumArray(int[] nums) {
int len = nums.length;
dp = new int[len+1];
for(int i = 0; i < len; i++)
dp[i+1] = dp[i] + nums[i];
}

public int sumRange(int i, int j) {
return dp[j+1] - dp[i];
}
}

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

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

}

public int sumRegion(int row1, int col1, int row2, int col2) {

return dp[row2+1][col2+1] + dp[row1][col1] - dp[row2+1][col1] - dp[row1][col2+1];

}
}