https://leetcode.com/problems/range-sum-query-immutable/description/
题目
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:
- You may assume that the array does not change.
- There are many calls to sumRange function.
题目翻译
题目解析
参考答案
#include <iostream>
#include <vector>
using namespace std;
class NumArray {
public:
NumArray(vector<int> nums) : m_Sum(nums.size()+1, 0) {
for (int i = 0; i < nums.size(); i++)
{
m_Sum[i + 1] = m_Sum[i] + nums[i];
}
}
int sumRange(int i, int j) {
return m_Sum[j + 1] - m_Sum[i];
}
private:
vector<int> m_Sum;
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
int main()
{
return 0;
}
class NumArray {
public:
vector<int> sums;
NumArray(vector<int> &nums) {
int i;
int iSize = nums.size();
sums = nums;
for (i = 1; i < iSize; i++)
{
sums[i] += sums[i-1];
}
}
int sumRange(int i, int j) {
printf("i = %d, j = %d, ", i, j);
if(i == 0)
{
printf("result = %d\n", sums[j]);
return sums[j];
}
else
{
printf("result = %d\n", sums[j] - sums[i-1]);
return sums[j] - sums[i-1];
}
}
};
// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);
