LeetCode 303. Range Sum Query

303. Range Sum Query - Immutable

一、问题描述

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

二、输入输出

Example:

Given nums = [-2, 0, 3, -5, 2, -1]sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3

三、解题思路

• 是按照DP TAG来选的题目，结果这道题我实在是看不出来DP的影子。DP是用来解决最优化问题的，这个题完全就没有最优化的影子。如果每次都从i加到j TLE错误。提示里说了会调用方法很多次。
• 全新的一个思路，只需要o(n)时间复杂度，o(n)的空间复杂度，如果不用sums数组，而是在原来的nums上修改，空间复杂度就成了o(1)
• 核心思想就是 sum(i, j) = sum(0, j) - sum(0, i-1) 就是把从0到i的和记录下来，之后让求从i到j的和，主要做一次减法就行了，只需要常数时间。

class NumArray {public:        vector<int> sums;        NumArray(vector<int> nums) {            if (nums.empty()) return;            sums.push_back(nums[0]);            for (int i = 1; i < nums.size(); ++i) {                int lastSum = sums[i-1];                sums.push_back(lastSum + nums[i]);            }        }        int sumRange(int i, int j) {            if(i == 0) return sums[j];            return (sums[j] - sums[i-1]);        }};/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */