303. Range Sum Query

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) -> 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.

链接： http://leetcode.com/problems/range-sum-query-immutable/

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。

public class NumArray {
    int[] partialSum;
    public NumArray(int[] nums) {
        partialSum = new int[nums.length + 1];
        for(int i = 1; i < partialSum.length; i++) {
            partialSum[i] = nums[i-1] + partialSum[i-1];
        }
    }
    
    public int sumRange(int i, int j) {
        return partialSum[j+1] - partialSum[i];
    }
}


/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */