Algorithm Practice
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304 Range Sum Query 2D - Immutable
1. Question
Given a 2D matrixmatrix, 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:
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Given matrix = [
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[3, 0, 1, 4, 2],
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[5, 6, 3, 2, 1],
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[1, 2, 0, 1, 5],
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[4, 1, 0, 1, 7],
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[1, 0, 3, 0, 5]
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]
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sumRegion(2, 1, 4, 3) -> 8
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sumRegion(1, 1, 2, 2) -> 11
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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 row 1 ≤ row2 and col1 ≤ col2.
2. Implementation
(1) DP
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class NumMatrix {
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int row;
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int col;
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int[][] sum;
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public NumMatrix(int[][] matrix) {
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
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return;
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}
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row = matrix.length;
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col = matrix[0].length;
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sum = new int[row + 1][col + 1];
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for (int i = 1; i <= row; i++) {
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for (int j = 1; j <= col; j++) {
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sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + matrix[i - 1][j - 1];
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}
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}
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}
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public int sumRegion(int row1, int col1, int row2, int col2) {
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return sum[row2 + 1][col2 + 1] - sum[row2 + 1][col1] - sum[row1][col2 + 1] + sum[row1][col1];
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}
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}
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/**
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* Your NumMatrix object will be instantiated and called as such:
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* NumMatrix obj = new NumMatrix(matrix);
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* int param_1 = obj.sumRegion(row1,col1,row2,col2);
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*/
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3. Time & Space Complexity
DP: 时间复杂度O(mn), 空间复杂度O(mn)