leetcode 数组不可变

最新推荐文章于 2025-12-18 17:00:31 发布

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#leetcode #算法 #职场和发展

本文介绍了一种使用前缀和数据结构来高效处理区间元素求和的算法。通过预先计算数组中每个位置到当前为止的元素总和，可以快速地在O(1)时间内回答任意区间内的元素和。具体实现包括构造函数和SumRange方法，后者返回指定索引范围内元素的总和。这种数据结构在处理大量查询时具有较高的效率。

Given an integer array nums, handle multiple queries of the following type:

Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.
Implement the NumArray class:

NumArray(int[] nums) Initializes the object with the integer array nums.
int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right])

type NumArray struct {
    sums []int
}

func Constructor(nums []int) NumArray {
    sums := make([]int, len(nums)+1)
    for i, v := range nums {
        sums[i+1] = sums[i] + v
    }
    return NumArray{sums}
}

func (na *NumArray) SumRange(i, j int) int {
    return na.sums[j+1] - na.sums[i]
}