[leetcode] 303. Range Sum Query - Immutable
2016-01-07 19:36
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Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
Example:
Note:
You may assume that the array does not change.
There are many calls to sumRange function.
这道题是查询区间段数据之和,题目难度为esay。
题目要求不改变原数组,同时提示会多次调用此函数,所以要尽可能降低函数的时间复杂度。循环加的方法时间复杂度跟区间长度有关,为了降低时间复杂度我们这里用空间换时间,把到每个下标为止的数据之和存下来,然后做减法就得到了区间段的数据之和,这里用到了动态规划的思想。具体代码:
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.
这道题是查询区间段数据之和,题目难度为esay。
题目要求不改变原数组,同时提示会多次调用此函数,所以要尽可能降低函数的时间复杂度。循环加的方法时间复杂度跟区间长度有关,为了降低时间复杂度我们这里用空间换时间,把到每个下标为止的数据之和存下来,然后做减法就得到了区间段的数据之和,这里用到了动态规划的思想。具体代码:
class NumArray { vector<int> curSum; public: NumArray(vector<int> &nums) { if(nums.size() > 0) { curSum.push_back(nums[0]); for(int i=1; i<nums.size(); i++) { curSum.push_back(curSum[i-1] + nums[i]); } } } int sumRange(int i, int j) { if((i > j) || (j >= curSum.size())) return 0; if(i==0) return curSum[j]; else return curSum[j]-curSum[i-1]; } };
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