https://leetcode.com/problems/range-sum-query-immutable/

用一個 sum 數(shù)組，sum[i] -- nums 中選出前 i 個元素，求和所得到的值。

這樣就有:

nums = [-2, 0, 3, -5, 2, -1] sum = [0, -2, -2, 1, -4, -2, -3]
sumRange(0, 2) -> sum[2 + 1] - sum[0] -> 1 sumRange(2, 5) -> sum[5 + 1] - sum[2] -> -1 sumRange(0, 5) -> sum[5 + 1] - sum[0] -> -3 public class NumArray {int[] sum;public NumArray(int[] nums) {sum = new int[nums.length + 1];for (int i = 0; i < nums.length; i++) {sum[i + 1] = sum[i] + nums[i]; } } public int sumRange(int i, int j) { return sum[j + 1] - sum[i]; } } /** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */

轉(zhuǎn)載于:https://www.cnblogs.com/chencode/p/range-sum-query-immutable.html

總結(jié)

以上是生活随笔為你收集整理的[LeetCode] 303. Range Sum Query - Immutable的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。