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

时间:2022-08-20 23:36:09

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);
*/