【动态规划】Leetcode编程题解:303. Range Sum Query
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题目:Given an integer array nums, find the sum of the elements between indicesi and j (i ≤ j), inclusive.
样例:
Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
提示:
- You may assume that the array does not change.
- There are many calls to sumRange function.
这道题目在数组较小的情况下可以直接将数组中的第i项加到第j项,但是如果数组规模较大,时间就会线性增加,所以可以定义一个数组,每个位置存储的是该位置之前所有的数之和,那么时间复杂度就可以降到常数级。
代码如下:
class NumArray {public: NumArray(vector<int> nums) { number.push_back(0); for(int i = 0; i < nums.size(); i++) { number.push_back(number.back() + nums[i]); } } int sumRange(int i, int j) { return number[j+1] - number[i]; }private: vector<int> number; };
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