标签：UNC between [] int ice numa cto any anti
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:
Approach #1: My code. [C++]
class NumArray {
public:
NumArray(vector<int> nums) {
nums_ = nums;
}
int sumRange(int i, int j) {
int ans = 0;
for (int k = i; k <= j; ++k)
ans += nums_[k];
return ans;
}
private:
vector<int> nums_;
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
Approach #2 (Caching) [Java]
private int[] sum;
public NumArray(int[] nums) {
sum = new int[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
sum[i + 1] = sum[i] + nums[i];
}
}
public int sumRange(int i, int j) {
return sum[j + 1] - sum[i];
}
303. Range Sum Query - Immutable
标签：UNC between [] int ice numa cto any anti
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10386163.html
