动态规划-二维数组区间求和

Description:

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).

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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

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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.

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问题描述:

在一个2维矩阵中，返回由左上角(row1, col1)和右下角(row2, col2)所固定的矩形中的元素的和。

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问题分析:

解法1
建立前缀数组，然后行遍历时累加

解法2
动态规划，可以看看这个链接:
https://leetcode.com/problems/range-sum-query-2d-immutable/discuss/75350/Clean-C++-Solution-and-Explaination-O(mn)-space-with-O(1)-time

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解法1

class NumMatrix {
    private int[][]dp;
    public NumMatrix(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return;

        int row = matrix.length,col = matrix[0].length;
        //dp中的每一行对应矩阵中每一行的前缀数组
        dp = new int[row][col + 1];
        for(int i = 0;i < row;i++){
            for(int j = 0;j < col;j++){
                dp[i][j + 1] = matrix[i][j] + dp[i][j];
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        int sum = 0;
        //遍历累加
        for(int i = row1;i <= row2;i++){
            sum += dp[i][col2 + 1] - dp[i][col1];
        }

        return sum;
    }
}

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解法2(动态规划):

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

        int m = matrix.length, 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][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + matrix[i - 1][j - 1];
            }
        }
    }

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