303. Range Sum Query - Immutable

问题：

给定一个数组，求任意区间[left, right]的元素和。

Example 1:
Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

Explanation
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))
 

Constraints:
1 <= nums.length <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= left <= right < nums.length
At most 10^4 calls will be made to sumRange.

　　

解法：前缀和（preSum）DP

• 在初始化数组时，构建前缀和数组，
  • pSum[i+1]：记录第 0～i 元素之和。
• base: pSum[0] = 0, pSum[1]=nums[0]
• 那么每次所求区间[left, right]
  • 即返回 pSum[right+1]-pSum[left]

 

代码参考：

class NumArray {
public:
    vector<int> pSum;
    NumArray(vector<int>& nums) {
        int n = nums.size();
        pSum.resize(n+1,0);
        for(int i=0; i<n; i++) {
            pSum[i+1] = pSum[i]+nums[i];
        }
    }
    
    int sumRange(int left, int right) {
        return pSum[right+1]-pSum[left];
    }
};

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