LeetCode Range Sum Query - Immutable
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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) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3
Note:
- You may assume that the array does not change
- There are many calls to sumRange function.
这个题的主要思路是为了防止频繁的sumRange操作导致大量的重复计算,因此在系统里面存上从0-j的元素的总和,这样在sumRange的时候可以直接按照位置关系相减即可,不用重复计算
class NumArray {public: //这个题的主要思路是为了防止频繁的sumRange操作导致大量的重复计算,因此在系统里面存上从0-j的元素的总和,这样在sumRange的时候可以直接按照位置关系相减即可,不用重复计算 vector<int> array; NumArray(vector<int> &nums) { for(int i=0;i<nums.size();i++) array.push_back(nums[i]); for(int i= 1;i<nums.size();i++) array[i] += array[i-1]; } int sumRange(int i, int j) { if(i<0||j>=array.size()||i>j) return 0; int total = 0; if(i==0) total = array[j]; else total = array[j] - array[i-1]; return total; }};// Your NumArray object will be instantiated and called as such:// NumArray numArray(nums);// numArray.sumRange(0, 1);// numArray.sumRange(1, 2);
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