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 ThreadTest{//求给定数组的[i,j]闭区间之内的元素的和
public static void main(String args[]){
int[] a={1,2,3,4,5};
sums(a);
System.out.println(sumsRange(3,4));
}
public static int [] sums;//这个里边的成员变量和成员函数貌似都必须定义为public static，不然不能用，具体不太懂
public static void sums(int[] nums){//定义了一个求数组nums和的数组sums
if(nums==null)
sums=null;
else if(nums.length==0)
sums=new int[0];
else{
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 static int sumsRange(int i,int j){//用上边的求和数组的差来求
if(sums==null)//这儿的判断依据是数组sums而不是nums
return 0;
if(i>j||i>sums.length||j<0)
return 0;
else if (i==0)
return sums[j];
else
return sums[j]-sums[i-1];
}
}