<LeetCode OJ> 303. Range Sum Query - Immutable

303. Range Sum Query - Immutable

My Submissions

Question

Total Accepted: 14338  Total Submissions: 59913  Difficulty: Easy

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:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

分析：

技巧题,，申请数组先算出结果，计算时直接获取结果进行相加减即可

class NumArray {
public:
    NumArray(vector<int> &nums)
    {//因为会被大量调用，所以先把所有的和算出来
        vec=nums;
        int sum = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            sum += nums[i];
            sumArray.push_back(sum);
        }
    }

    int sumRange(int i, int j) {
       int sumi=sumArray[i];
       int sumj=sumArray[j];
       return sumj-sumi+vec[i];
    }
    vector<int> vec;
    vector<int> sumArray;
};


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

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.youkuaiyun.com/ebowtang/article/details/50354078

原作者博客：http://blog.youkuaiyun.com/ebowtang