303. Range Sum Query - Immutable

问题描述

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.

题目链接：

---

思路分析

写一个类，来计算一个数组从i到j的所有元素的和。

用一个向量保存每一个从0到这个元素的和，计算时从sum里面减就可以。sum[j] - sum[i-1]

代码

class NumArray {
public:
    vector<int> sum;
    NumArray(vector<int> nums) {
        int temp = 0;
        for (int i = 0; i < nums.size(); i++){
            temp += nums[i];
            sum.push_back(temp);
        }
    }

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

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

时间复杂度：$O(1)$
空间复杂度：$O(n)$

---

反思

一开始的方法是保存数组，然后加和计算的，结果TLE了

class NumArray {
public:
    vector<int> mynums;
    NumArray(vector<int> nums) {
        for (int i = 0; i < nums.size(); i++){
            mynums.push_back(nums[i]);
        }
    }

    int sumRange(int i, int j) {
        int result = 0;
        for (int k = i; k <= j; k++){
            result += mynums[k];
        }
        return result;
    }
};

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

学到一招：的用法，相当于从这个数组里面自动取一个元素出来，简洁。

class NumArray {
public:
    vector<int> sum;
    NumArray(vector<int> nums) {
        int temp = 0;
        for (int i:nums){
            temp += i;
            sum.push_back(temp);
        }
    }

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