Leetcode: Range Sum Query - Immutable

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

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.

 第二遍方法（更好）：定义prefix sum array length: nums.length+1

public class NumArray {
    int[] arr;

    public NumArray(int[] nums) {
        if (nums.length == 0) return;
        arr = new int[nums.length+1];
        for (int i=0; i<nums.length; i++) {
            arr[i+1] = arr[i] + nums[i];
        }
    }

    public int sumRange(int i, int j) {
        return arr[j+1]-arr[i];
    }
}


// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);

第一遍方法：定义prefix sum length 为nums.length, 这样sumRange要讨论i=0的情况

public class NumArray {
    int[] leftSums;

    public NumArray(int[] nums) {
        leftSums = new int[nums.length];
        if (nums.length == 0) return;
        leftSums[0] = nums[0];
        for (int i=1; i<nums.length; i++) {
            leftSums[i] = leftSums[i-1] + nums[i];
        }
    }

    public int sumRange(int i, int j) {
        return i==0? leftSums[j] : leftSums[j] - leftSums[i-1];
    }
}


// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);