Range Sum Query - Immutable (E)

题目

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:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

题意

给定一个数组，求从下标i到j的元素之和。

思路

因为方法可能被多次调用，因而可以用缓存进行优化：另设数组sum，sum[i]表示nums中前i个元素之和。因此i到j的元素之和就可以表示为sum[j + 1] - sum[i]。

代码实现

Java

class NumArray {
    private int[] sum;

    public NumArray(int[] nums) {
        sum = new int[nums.length + 1];
        for (int i = 0; i < nums.length; i++) {
            sum[i + 1] = sum[i] + nums[i];
        }
    }

    public int sumRange(int i, int j) {
        return sum[j + 1] - sum[i];
    }
}

0303. Range Sum Query - Immutable (E)

原文：https://www.cnblogs.com/mapoos/p/13197163.html