[LeetCode] 303. Area and Search-Array Immutable (C++)

1 topic description

Given an integer array nums, find the sum of the elements in the range from index i to j (i ≤ j), including two points i and j.
Implement the NumArray class:

• NumArray(int[] nums) uses the array nums to initialize the object
• int sumRange(int i, int j) returns the sum of the elements in the array nums from index i to j (i ≤ j), including two points i and j (that is, sum(nums[i], nums[i + 1] ,…, Nums[j]))

2 Example description

输入：
[“NumArray”, “sumRange”, “sumRange”, “sumRange”]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出：
[null, 1, -1, -3]
解释：
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

3 Problem solving tips

0 <= nums.length <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= i <= j < nums.length
最多调用 10^4 次 sumRange 方法

4 Problem-solving ideas

Define the sum of the array from nums[0] to nums[i], but the sum of i to j must be calculated from i+1, otherwise it will fall to the last one.

5 Detailed source code (C++)

class NumArray {
    
    
public:
    vector<int> sum ; //定义从nums[0]到nums[i]的数组和，但是计算i到j的和要从i+1开始算，否则会落下最后一个
    NumArray(vector<int>& nums) {
    
    
        sum.resize( nums.size() + 1 ) ;
        for ( int i = 0 ; i < nums.size() ; i ++ )
        {
    
    
            sum[i + 1] = sum[i] + nums[i] ;
        }
    }
    
    int sumRange(int i, int j) {
    
    
        return sum[j + 1] - sum[i] ;
    }
};

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