303. Range Sum Query - Immutable(动态规划）

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：
由于开始是写PAT现在写leetcode提交方式很多不一样，这题提交方式看得我很懵逼啊，于是搜了一下什么意思；这道题首先求和了从0到i(0<=i<=n),然后如果求sum(i,j),如果i==0,
则输出v[j],如果i!=0，输出v[j]-v[i-1];

class NumArray {
private:
    vector<int> v;
public:
    NumArray(vector<int> nums) {
        if(nums.size()==0)
            return ;
        v.push_back(nums[0]);
        for(int i=1;i<nums.size();i++)
            v.push_back(v[i-1]+nums[i]);
    }
    int sumRange(int i, int j) {
        if(i==0)
            return v[j];
        return v[j]-v[i-1];
    }
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

原文地址：https://www.cnblogs.com/A-Little-Nut/p/8322054.html