区域和检索 - 数组不可变

给定一个整数数组 nums，求出数组从索引 i 到 j (i ≤ j) 范围内元素的总和，包含 i, j 两点。

示例：

给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

普通解法：

class NumArray {
private:
    vector<int> dp;
public:
    NumArray(vector<int>& nums) {
        dp = nums;
    }
    
    int sumRange(int i, int j) {
        int num = 0;
        while(i<=j)
        {
            num += dp[i];
            ++i;
        }
        return num;
    }
};

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

巧妙解法：

class NumArray {
public:
    NumArray(vector<int>& nums) {
       if(!nums.empty())
       {
            _arr.push_back(nums[0]);
            for(int i = 1;i<nums.size();++i)
            {
                _arr.push_back(_arr.back()+nums[i]);
            }
       }
    }
    
    int sumRange(int i, int j) {
        if(i == 0)
        {
            return _arr[j];
        }
        return _arr[j]-_arr[i-1];
    }
public:
    vector<int> _arr;
};

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

标签：检索,vector,NumArray,int,nums,区域,数组,sumRange
来源： https://blog.csdn.net/C1029323236/article/details/89077319