303 Range Sum Query - Immutable 区域和检索 - 数组不可变
Description:
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.
题目描述:
给定一个整数数组 nums，求出数组从索引 i 到 j (i ≤ j) 范围内元素的总和，包含 i, j 两点。
示例 :
给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
说明:
你可以假设数组不可变。
会多次调用 sumRange 方法。
思路:
动态规划(前缀和), 记录下从 0开始到 i的数组之和
调用时返回 sum_[j] - sum_[i] + nums[i]即可
初始化类时间复杂度O(n), 空间复杂度O(n)
查询时间复杂度O(1), 空间复杂度O(1)
代码:
C++:
class NumArray
{
private:
vector<int> nums;
vector<int> sum_;
public:
NumArray(vector<int>& nums)
{
this -> nums = nums;
int temp = 0;
for (int i = 0; i < nums.size(); i++)
{
temp += nums[i];
sum_.push_back(temp);
}
}
int sumRange(int i, int j)
{
return sum_[j] - sum_[i] + nums[i];
}
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray* obj = new NumArray(nums);
* int param_1 = obj->sumRange(i,j);
*/
Java:
class NumArray {
private int[] sum_;
private int[] nums;
public NumArray(int[] nums) {
sum_ = new int[nums.length];
int temp = 0;
for (int i = 0; i < nums.length; i++) {
temp += nums[i];
sum_[i] = temp;
}
this.nums = nums;
}
public int sumRange(int i, int j) {
return sum_[j] - sum_[i] + nums[i];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
Python:
class NumArray:
def __init__(self, nums: List[int]):
self.nums = nums
temp = 0
sum_ = []
for i in nums:
temp += i
sum_.append(temp)
self.sum_ = sum_
def sumRange(self, i: int, j: int) -> int:
return self.sum_[j] - self.sum_[i] + self.nums[i]
# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)
