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;
}

相关题目

results matching ""

    No results matching ""