[LeetCode]303. Range Sum Query - Immutable

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

多次计算[i, j]数组和

空间复杂度O(1)即可，dp保存[0, i]的和。此处又用到dp多开一位的trick

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