leetcode 303. Range Sum Query

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.

解法1（self）：

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

解法2：子数列和，动态规划

因为是大量调用计算，需要考虑计算效率。这里使用子数列和的方法（求得子数列的累加）；

class NumArray {public:    NumArray(const vector<int> &nums) {        accu.push_back(0);        for (const int num : nums)            accu.push_back(accu.back() + num);    }    int sumRange(int i, int j) {        return accu[j + 1] - accu[i];    }    vector<int> accu;};// Your NumArray object will be instantiated and called as such:// NumArray numArray(nums);// numArray.sumRange(0, 1);// numArray.sumRange(1, 2);