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:

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.

303. Range Sum Query - Immutable 的變形，這題是2D陣列，給左上角和右下角的點，這兩點的行和列組成了一個矩形，求這個矩形裡所有數字的和。

解法：DP， 建立一個二維陣列dp，其中dp[i][j]表示累計區間(0, 0)到(i, j)這個矩形區間所有數字的和，求(r1, c1)到(r2, c2)的矩形區間和時，只需dp[r2][c2] - dp[r2][c1 - 1] - dp[r1 - 1][c2] + dp[r1 - 1][c1 - 1]即可。

Java:

private int[][] dp;

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

public int sumRegion(int row1, int col1, int row2, int col2) {
    int iMin = Math.min(row1, row2);
    int iMax = Math.max(row1, row2);
    
    int jMin = Math.min(col1, col2);
    int jMax = Math.max(col1, col2);
    
    return dp[iMax + 1][jMax + 1] - dp[iMax + 1][jMin] - dp[iMin][jMax + 1] + dp[iMin][jMin];    
}

Python:

class NumMatrix(object):
      def __init__(self, matrix):
          if matrix is None or not matrix:
              return
          n, m = len(matrix), len(matrix[0])
          self.sums = [ [0 for j in xrange(m+1)] for i in xrange(n+1) ]
          for i in xrange(1, n+1):
              for j in xrange(1, m+1):
                  self.sums[i][j] = matrix[i-1][j-1] + self.sums[i][j-1] + self.sums[i-1][j] - self.sums[i-1][j-1]
    

      def sumRegion(self, row1, col1, row2, col2):
          row1, col1, row2, col2 = row1+1, col1+1, row2+1, col2+1
          return self.sums[row2][col2] - self.sums[row2][col1-1] - self.sums[row1-1][col2] + self.sums[row1-1][col1-1]

Python:　　

# Time:  ctor:   O(m * n),
#        lookup: O(1)
# Space: O(m * n)

class NumMatrix(object):
    def __init__(self, matrix):
        """
        initialize your data structure here.
        :type matrix: List[List[int]]
        """
        if not matrix:
            return

        m, n = len(matrix), len(matrix[0])
        self.__sums = [[0 for _ in xrange(n+1)] for _ in xrange(m+1)]
        for i in xrange(1, m+1):
            for j in xrange(1, n+1):
                self.__sums[i][j] = self.__sums[i][j-1] + matrix[i-1][j-1]
        for j in xrange(1, n+1):
            for i in xrange(1, m+1):
                self.__sums[i][j] += self.__sums[i-1][j]

    def sumRegion(self, row1, col1, row2, col2):
        """
        sum of elements matrix[(row1,col1)..(row2,col2)], inclusive.
        :type row1: int
        :type col1: int
        :type row2: int
        :type col2: int
        :rtype: int
        """
        return self.__sums[row2+1][col2+1] - self.__sums[row2+1][col1] - \
               self.__sums[row1][col2+1] + self.__sums[row1][col1]　

C++:

class NumMatrix {
private:
    int row, col;
    vector<vector<int>> sums;
public:
    NumMatrix(vector<vector<int>> &matrix) {
        row = matrix.size();
        col = row>0 ? matrix[0].size() : 0;
        sums = vector<vector<int>>(row+1, vector<int>(col+1, 0));
        for(int i=1; i<=row; i++) {
            for(int j=1; j<=col; j++) {
                sums[i][j] = matrix[i-1][j-1] + 
                             sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] ;
            }
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        return sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1];
    }
};

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[LeetCode] 303. Range Sum Query - Immutable 區域和檢索 - 不可變

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