LeetCode|LeetCode #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
【LeetCode|LeetCode #303 Range Sum Query - Immutable 区域和检索 - 数组不可变】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, 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) - 线段树 Segment_tree-wiki
研究中, 将完成于LeetCode #307. Range Sum Query - Mutable 区域和检索 - 数组可修改
C++:
class NumArray {
private:
vector nums;
vector sum_;
public:
NumArray(vector& 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)
推荐阅读
- 【Leetcode/Python】001-Two|【Leetcode/Python】001-Two Sum
- leetcode|leetcode 92. 反转链表 II
- 二叉树路径节点关键值和等于目标值(LeetCode--112&LeetCode--113)
- LeetCode算法题-11.|LeetCode算法题-11. 盛最多水的容器(Swift)
- LeetCode(03)Longest|LeetCode(03)Longest Substring Without Repeating Characters
- Leetcode|Leetcode No.198打家劫舍
- [leetcode数组系列]1两数之和
- 数据结构和算法|LeetCode 的正确使用方式
- leetcode|今天开始记录自己的力扣之路
- LeetCode|LeetCode 876. 链表的中间结点