303. Range Sum Query - Immutable [Leetcode］

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

Note:

1. You may assume that the array does not change.
2. 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);