Leetcode[303, 304] Range Sum Query - Immutable
07 Dec 2015###Task1 Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
Example: Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3 Note: You may assume that the array does not change. There are many calls to sumRange function.
###Java ####O(n)
public class NumArray {
int[] sums;
public NumArray(int[] nums) {
sums = new int[nums.length + 1];
int sum = 0;
sums[0] = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
sums[i + 1] = sum;
}
}
public int sumRange(int i, int j) {
return sums[j + 1] - sums[i];
}
}
// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);
###Points
计算辅助数组sums: sums[0] = 0, sums[i+1] = sums[i] + nums[i]
###Task2 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:
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 Note:
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.
###Java
public class NumMatrix {
int[][] sums;
public NumMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return;
}
sums = new int[matrix.length + 1][matrix[0].length + 1];
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
sums[i + 1][j + 1] = matrix[i][j] + sums[i + 1][j] + sums[i][j + 1] - sums[i][j] ;
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
if (sums == null) {
return 0;
}
return sums[row2 + 1][col2 + 1] - sums[row1][col2 + 1] - sums[row2 + 1][col1] + sums[row1][col1];
}
}
###Points
- Sums always larger than the original: why? handle corner case.