Range Sum Query 2D - Immutable
Desicription
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).
Range Sum Query 2D The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
Example:
代码语言：javascript复制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) -> 12Note:
- 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.
Solution
代码语言：javascript复制class NumMatrix {
vector<vector<int>> dp;
public:
explicit NumMatrix(const vector<vector<int>> &matrix) {
if(matrix.empty()) {
return ;
}
dp = vector<vector<int>>(matrix.size(), vector<int>(matrix[0].size()));
unsigned long row = matrix.size();
unsigned long col = matrix[0].size();
dp[0][0] = matrix[0][0];
for(int i = 1; i < row; i ) {
dp[i][0] = dp[i - 1][0] matrix[i][0];
}
for(int i = 1; i < col; i ) {
dp[0][i] = dp[0][i - 1] matrix[0][i];
}
for(int i = 1; i < row; i ) {
for(int j = 1; j < col; j ) {
dp[i][j] = dp[i - 1][j] dp[i][j - 1] - dp[i - 1][j - 1] matrix[i][j];
}
}
}
int sumRegion(int row1, int col1, int row2, int col2) {
if(dp.empty()) {
return 0;
}
if(row1 == 0 && col1 == 0) {
return dp[row2][col2];
} else if(row1 == 0) {
return dp[row2][col2] - dp[row2][col1 - 1];
} else if(col1 == 0) {
return dp[row2][col2] - dp[row1 - 1][col2];
} else {
return dp[row2][col2] - dp[row2][col1 - 1] - dp[row1 - 1][col2] dp[row1 - 1][col1 - 1];
}
}
};
