题目：

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) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3

 

Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

链接： 

3/7/2017

看别人答案的，之后就算有了思路还是做错。原因：没有仔细想清楚每个变量和数组的意义，比如start/end的元素是否包括，辅助数组的和是否包括当前值，以及辅助数组的长度。

1 public class NumArray { 2     int[] partialSum; 3     public NumArray(int[] nums) { 4         partialSum = new int[nums.length + 1]; 5         for(int i = 1; i < partialSum.length; i++) { 6             partialSum[i] = nums[i-1] + partialSum[i-1]; 7         } 8     } 9     10     public int sumRange(int i, int j) {11         return partialSum[j+1] - partialSum[i];12     }13 }14 15 16 /**17  * Your NumArray object will be instantiated and called as such:18  * NumArray obj = new NumArray(nums);19  * int param_1 = obj.sumRange(i,j);20  */