Given an integer array nums, find the sum of the elements between
indices i and j (i ≤ j), inclusive.數組Example: Given nums = [-2, 0, 3, -5, 2, -1]code
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.element
思路: 求區間的和就想到前n項和,即sumRange(i, j) = sum[j + 1] - sum[i]. 爲了方便, 前n項和數組通常比原數組開闢的多一個, 由於前0項和爲0. io
時間複雜度: O(n)
空間複雜度: O(n)table
public class NumArray { private int[] sum; public NumArray(int[] nums) { sum = new int[nums.length + 1]; sum[0] = 0; for (int i = 1; i <= nums.length; i++) { sum[i] = sum[i - 1] + nums[i - 1]; } } public int sumRange(int i, int j) { return sum[j + 1] - sum[i]; } } // Your NumArray object will be instantiated and called as such: // NumArray numArray = new NumArray(nums); // numArray.sumRange(0, 1); // numArray.sumRange(1, 2);