range-sum-query-immutable

Range Sum Query - Immutable – LeetCode 303

Problem

Description

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

Answer

Original

Code

class NumArray {
private:
    vector<int> map{0};
public:
    NumArray(vector<int> nums) {
        for(int i = 0, acc = 0; i != nums.size(); ++i){
            acc += nums[i];
            map.push_back(acc);
        }
    }
    
    int sumRange(int i, int j) {
        return map[j+1] - map[i];
    }
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

思路

因为不可变所以建立一个缓存，用空间换时间，对于map[n] = num[1] + … + num[n]，然后自然可得sumRange(i,j) = map[j] - map[i]。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$163$ ms，排名$16.83\%$

Better

思路

还没看到更好的思路。