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:

1. You may assume that the array does not change.

2. There are many calls to sumRange function.

Analysis

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

重点在于sumRange() 需要较高的效率，比如O(1)的时间复杂度。因此在初始化时要进行处理，最直觉的就是关于range sum，可以新建数组存入[0, i)的sums，这样需要[i, j]的时候，sums[j + 1] - sums[i]即可。小技巧是生成nums.length + 1的sums[]，而sums[i]代表了[0, i)，也就是不包含i的和。

Solution

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

    public int sumRange(int i, int j) {
        if (i == 0) return sums[j];
        return sums[j] - sums[i - 1];
    }
}