303. Range Sum Query - Immutable

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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.

给定一个int数组，返回i和j之间的元素总和，注意时间复杂度

思路：dp问题，  sums[i]表示0到i之间数据的总和，这样sums[j]-sums[i]就是i到j之间的总和

 1 class NumArray {
 2     
 3     int[] sums;
 4     public NumArray(int[] nums) {
 5         sums = new int[nums.length];
 6         if (nums.length == 0) return;
 7         sums[0] = nums[0];
 8         for (int i=1;i<nums.length;i++)
 9         {
10             sums[i] = sums[i-1] + nums[i];
11         }        
12     }
13     
14     public int sumRange(int i, int j) {
15       return i==0?sums[j]:sums[j]-sums[i-1];  
16     }
17 }

303. Range Sum Query - Immutable

标签：tab   数据   col   man   hat   array   []   div   amp   

原文地址：http://www.cnblogs.com/wzj4858/p/7687805.html