LeetCode Top 100 Liked Questions 308. Range Sum Query 2D - Mutable (Java版; Hard)

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LeetCode Top 100 Liked Questions 308. Range Sum Query 2D - Mutable (Java版; Hard)

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
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10
Note:
The matrix is only modifiable by the update function.
You may assume the number of calls to update and sumRegion function is distributed evenly.
You may assume that row1 ≤ row2 and col1 ≤ col2.

第一次做; 没有使用线段树或者树状数组, 而是多次使用了一维前缀和

class NumMatrix {
    
    private int[][] matrix;
    private int[][] sum;
    
    public NumMatrix(int[][] matrix) {
        this.matrix = matrix;
        getSumArr(matrix);
    }
    
    public void update(int row, int col, int val) {
        int diff = val - matrix[row][col];
        //记得更新matrix[row][col], 要不多次修改同一个位置后, 还是用最原始的matrix计算diff的值, 这样就错了!
        matrix[row][col] = val;
        //细节: j的起始值, 要时刻牢记sum[i][j]的含义
        for(int j=col+1; j<sum[0].length; j++){
            sum[row][j] = sum[row][j] + diff;
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        int res = 0;
        for(int i=row1; i<=row2; i++){
            res = res + sum[i][col2+1] - sum[i][col1];
        }
        return res;
    }
    
    private void getSumArr(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matrix索引为i那一行的前j个数的和
        sum[i][j]=matrix[i][0]+matrix[i][1]+...+matrix[i][j-1]
        sum[i][j]=sum[i][j-1] + matrix[i][j-1]
        */
        sum = new int[rows][cols+1];
        for(int i=0; i<rows; i++){
            for(int j=1; j<=cols; j++){
                sum[i][j] = sum[i][j-1] + matrix[i][j-1];
            }
        }
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * obj.update(row,col,val);
 * int param_2 = obj.sumRegion(row1,col1,row2,col2);
 */

力扣优秀题解(https://leetcode-cn.com/problems/range-sum-query-2d-mutable/solution/wu-xu-xian-duan-shu-de-gao-xiao-jie-fa-20ms-ji-bai/)

新建一个数组rowSumArr来存放各个位置(i, j) 的前j+1项的和，即从(i, 0) 到 (i, j)各个元素的和。
若更新（执行update方法）的时候，只需要更新要被更新位置(i, j)所在行的后面的元素的rowSumArr的值即可。
若要计算二维区域和（执行sumRegion方法）的时候，只要求得二维区域每一行的元素之和，之后累加即可。
这里刚好可以利用到前面保存的rowSumArr上的值，每一行的和只要O(1)的时间复杂度即可，整体只要O(n)时间复杂度。

class NumMatrix {

        private int[][] matrix;
        private int[][] rowSumArr;  // 保存每个元素(i, j)在第i行的前j+1项的和。
        private int rowCount;
        private int colCount;

        public NumMatrix(int[][] matrix) {
            this.matrix = matrix;
            if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
                return;
            }

            rowCount = matrix.length;
            colCount = matrix[0].length;
            rowSumArr = new int[rowCount][colCount];
            for (int i = 0; i < rowCount; i++) {
                rowSumArr[i][0] = matrix[i][0];
                for (int j = 1; j < colCount; j++) {
                    rowSumArr[i][j] = rowSumArr[i][j-1] + matrix[i][j];
                }
            }
        }

        public void update(int row, int col, int val) {
            matrix[row][col] = val;
            int fromCol = col;
            if (col == 0) {
                rowSumArr[row][col] = matrix[row][col];
                fromCol = col + 1;
            }
            for (int j = fromCol; j < colCount; j++) {
                rowSumArr[row][j] = rowSumArr[row][j-1] + matrix[row][j];
            }
        }

        public int sumRegion(int row1, int col1, int row2, int col2) {
            int sum = 0;
            for (int i = row1; i <= row2; i++) {
                sum += col1 == 0 ? rowSumArr[i][col2] : rowSumArr[i][col2] - rowSumArr[i][col1-1];
            }
            return sum;
        }
    }

树状数组的优秀讲解