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:

You may assume that the array does not change.
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。

 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  */

转载于:https://www.cnblogs.com/panini/p/6517670.html

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