題目：

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.

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

3/7/2017

看別人答案的，之后就算有了思路還是做錯(cuò)。原因：沒(méi)有仔細(xì)想清楚每個(gè)變量和數(shù)組的意義，比如start/end的元素是否包括，輔助數(shù)組的和是否包括當(dāng)前值，以及輔助數(shù)組的長(zhǎng)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 */

?

轉(zhuǎn)載于:https://www.cnblogs.com/panini/p/6517670.html

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