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

法一：暴力

 1 class NumArray {
 2
 3 public:
 4     vector<int> Nums;
 5
 6     NumArray(vector<int> nums) {
 7         Nums =nums;
 8     }
 9
10     int sumRange(int i, int j) {
11         int res = 0;
12         for(int k = i;k<=j;k++)
13             res+=Nums[k];
14         return res;
15     }
16 };

法二：

sumRange(i,j)=sum[j+1]−sum[i]

 1 class NumArray {
 2
 3 public:
 4     vector<int> Nums;
 5     vector<int> Sums;
 6
 7
 8     NumArray(vector<int> nums) {
 9         Nums =nums;
10         Sums = vector<int>(nums.size()+1,0);
11         for(int i = 0;i<nums.size();i++)
12             Sums[i+1]=Sums[i]+nums[i];
13     }
14
15     int sumRange(int i, int j) {
16         int res = 0;
17         res = Sums[j+1]-Sums[i];
18         return res;
19     }
20 };

原文地址：https://www.cnblogs.com/zle1992/p/10466479.html