303. Range Sum Query - Immutable [Leetcode］

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本文介绍了一种使用前缀和数组优化区间和查询的方法。通过预处理整数数组的累积和，可以快速计算任意两个索引之间的元素总和，适用于数组不变而查询频繁的场景。

Problem

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.

Solution

想到用另外一个同样大小的数组 sumArr[N] , 每个元素 summArr[ i ] 表示从 0 到 i （inclusive）的和。

这样求 i 到 j 的和，就是 sumArr[ j ] - sumsArr[ i -1]

class NumArray {
    vector<int> sumArr;
public:
    NumArray(vector<int> &nums)  {
        const int N = nums.size();
        sumArr.resize( N, 0);
        if(N > 0){
            sumArr[0] = nums[0];
            for( int i = 1; i < N; i++){
                sumArr[i] = sumArr[i-1] + nums[i];
            }
        }
    }

    int sumRange(int i, int j) {
        if( i == 0) return sumArr[j];
        return sumArr[j] - sumArr[i - 1];
    }
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);