[LeetCode] 304. Range Sum Query 2D - Immutable 二维区域和检索 - 不可变

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:

  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 的变形,这题是2D数组,给左上角和右下角的点,这两点的行和列组成了一个矩形,求这个矩形里所有数字的和。

解法:DP, 建立一个二维数组dp,其中dp[i][j]表示累计区间(0, 0)到(i, j)这个矩形区间所有数字的和,求(r1, c1)到(r2, c2)的矩形区间和时,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 区域和检索 - 不可变

 

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posted @ 2018-10-18 09:31  轻风舞动  阅读(330)  评论(0编辑  收藏  举报