LeetCode 304. Range Sum Query 2D - Immutable (Java版; Medium)

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LeetCode Top 100 Liked Questions 304. Range Sum Query 2D - Immutable (Java版; Medium)

题目描述

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:
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.

第一次做; 核心:1)sum[i][j]表示(0,0),(i-1,j-1)这个矩形范围内的元素求和 2)行数和列数比原始的matrix加一是为了不用检查数组索引是否越界, 因为数组索引会涉及减1操作 3)sum[][]的建立过程和sumRegion()中对sum[][]的使用是不同的, 见题解图示

class NumMatrix {
    private int[][] sum;
    
    public NumMatrix(int[][] matrix) {
        getSum(matrix);
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        //
        if(sum==null)
            return 0;
        //核心: 不要跟矩形面积搞混了, 边界处要想清楚, sum[i][j]表示(0,0),(i-1,j-1)这个矩形范围内的元素求和, 包括边界,
        //所以得是减sum[row2+1][col1],而不能减不能减sum[row2+1][col1+1], 这样会把边界上的值减掉! 就错了
        return sum[row2+1][col2+1] - sum[row2+1][col1] - sum[row1][col2+1] + sum[row1][col1];
        
    }
    
    private void getSum(int[][] matrix){
        //input check
        if(matrix==null || matrix.length==0 || matrix[0]==null || matrix[0].length==0)
            return;
        int rows = matrix.length, cols = matrix[0].length;
        //sum[i][j]表示(0,0),(i-1,j-1)这个矩形范围内的元素求和
        //sums[i+1][j+1] represents the sum of area from matrix[0][0] to matrix[i][j]
        sum = new int[rows+1][cols+1]; //行数和列数加一是为了不用检查数组索引是否越界, 因为数组索引会涉及减1操作
        for(int i=1; i<=rows; i++){
            for(int j=1; j<=cols; j++){
                //下面使用的索引涉及减1操作, 所以i和j需要从1开始
                sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + matrix[i-1][j-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);
 */

LeetCode优秀题解 用C++写的, 思想是相同的, 精髓是图示

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];
    }
};