303. 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.

public class NumArray {
	private static int[] nums;
	private static int[] sums;
    public NumArray(int[] nums) {
        if(nums==null)
			return ;
		this.nums = nums;
		sums=new int[nums.length];//注意为什么不nums=sums,这样的话跟着变，这样的话原始数组也会变，这样不科学
		if (nums.length == 1) {
			sums[0] = nums[0];
			return;
		}
		for (int x = 1; x < nums.length; x++) {
			sums[x] = sums[x - 1] + nums[x];
		}
    }

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


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