Leetcode:303. Range Sum Query

Description:

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

解题思路

本题就是求区间内的所有数字之和，数组的数字及其下标不会改变。一开始会习惯性的采用传统的遍历相加求解，但该算法无法通过Leetcode；因此此时就要考虑重新设一个数组sum用于累积数字之和，其中sum[i+1]表示【0，i】区间的数字之和，那么【i，j】就可表示为sum[j+1]-sum[i]，应注意的是i =0 时sum【i】= 0

代码

class NumArray {public:    NumArray(vector<int> nums) {        sum.push_back(0);        for (int i = 0; i < nums.size();i++) {            sum.push_back(sum.back()+nums[i]);        }    }    int sumRange(int i, int j) {        return sum[j+1]-sum[i];    }    vector<int> sum;};/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j);