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]

  1. sumRange(0, 2) -> 1
  2. sumRange(2, 5) -> -1
  3. 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

  1. // Range Sum Query - Immutable
  2. public class NumArray {
  3.     // Time Complexity: O(n), Space Complexity: O(1)
  4.     public NumArray(int[] nums) {
  5.         this.f = new int[nums.length];
  6.         int sum = 0;
  7.         for (int i = 0; i < nums.length; ++i) {
  8.             sum += nums[i];
  9.             f[i] = sum;
  10.         }
  11.     }
  13.     // Time Complexity: O(1), Space Complexity: O(1)
  14.     public int sumRange(int i, int j) {
  15.         return f[j] - (i == 0 ? 0 : f[i - 1]);
  16.     }
  17.     private final int[] f;
  18. }

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/dp/range-sum-query-immutable.html