题目描述
给定一个二维矩阵,计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2)。
示例:
给定 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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/range-sum-query-2d-immutable
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C++
暴力超时非得动态规划
class NumMatrix {
public:
vector<vector<int>> sums;
//动态规划
//表中递归关系式f(i,j)为以i,j为右下角元素的矩阵的元素和
//f(i,j)=f(i,j-1)+f(i-1,j)-f(i-1,j-1)+matrix[i][j]
NumMatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
if (m > 0) {
int n = matrix[0].size();
sums.resize(m + 1, vector<int>(n + 1));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sums[i + 1][j + 1] = sums[i][j + 1] + sums[i + 1][j] - sums[i][j] + matrix[i][j];
}
}
}
}
int sumRegion(int row1, int col1, int row2, int col2) {
return sums[row2 + 1][col2 + 1] - sums[row1][col2 + 1] - sums[row2 + 1][col1] + sums[row1][col1];
}
};