Range Sum Query - Immutable
描述
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.
分析
令状态f[i]为0到i元素之间的和，则状态转移方程为 f[i] = f[i-1] + nums[i]。f[i]本质上是累加和，有了f[i]，则范围[i,j]之间的和等于f[j] - f[i-1]。
代码
// Range Sum Query - Immutable
public class NumArray {
// Time Complexity: O(n), Space Complexity: O(1)
public NumArray(int[] nums) {
this.f = new int[nums.length];
int sum = 0;
for (int i = 0; i < nums.length; ++i) {
sum += nums[i];
f[i] = sum;
}
}
// Time Complexity: O(1), Space Complexity: O(1)
public int sumRange(int i, int j) {
return f[j] - (i == 0 ? 0 : f[i - 1]);
}
private final int[] f;
}