二维子矩阵的和——二维前缀和

代码1：

一维前缀和

1. 记录每一行的前缀和，在求值时将各行相加

type NumMatrix struct {
    sums [][]int
}

func Constructor(matrix [][]int) NumMatrix {
    sums := make([][]int, len(matrix))
    for i, row := range matrix {
        sums[i] = make([]int, len(row)+1)
        for j, v := range row {
            sums[i][j+1] = sums[i][j] + v
        }
    }
    return NumMatrix{sums}
}

func (nm *NumMatrix) SumRegion(row1, col1, row2, col2 int) (sum int) {
    for i := row1; i <= row2; i++ {
        sum += nm.sums[i][col2+1] - nm.sums[i][col1]
    }
    return
}

代码2：

二维前缀和

1. 直接得到当前节点对于（0，0）位置所包裹的矩阵和，在求值时进行运算，即可得到答案

type NumMatrix struct {
    sums [][]int
}

func Constructor(matrix [][]int) NumMatrix {
    m := len(matrix)
    if m == 0 {
        return NumMatrix{}
    }
    n := len(matrix[0])
    sums := make([][]int, m+1)
    sums[0] = make([]int, n+1)
    for i, row := range matrix {
        sums[i+1] = make([]int, n+1)
        for j, v := range row {
            sums[i+1][j+1] = sums[i+1][j] + sums[i][j+1] - sums[i][j] + v
        }
    }
    return NumMatrix{sums}
}

func (nm *NumMatrix) SumRegion(row1, col1, row2, col2 int) int {
    return nm.sums[row2+1][col2+1] - nm.sums[row1][col2+1] - nm.sums[row2+1][col1] + nm.sums[row1][col1]
}