# LeetCode 303. Range Sum Query - Immutable

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.

dp[i] 表示 nums[0]到nums[i-1]的和，如此做是为了求sumRange时不需要额外讨论 i == 0的情况。

Time Complexity: NumArray, O(n). sumRange, O(1). n = nums.length.

Space: O(n).

AC Java:

 1 public class NumArray {
2     private int [] dp;
3     public NumArray(int[] nums) {
4         dp = new int[nums.length+1]; //dp[i]表示从nums[0]到nums[i-1]的和
5         for(int i = 1; i<=nums.length; i++){
6             dp[i] = dp[i-1] + nums[i-1];
7         }
8     }
9
10     public int sumRange(int i, int j) {
11         return dp[j+1] - dp[i];
12     }
13 }
14
15 // Your NumArray object will be instantiated and called as such:
16 // NumArray numArray = new NumArray(nums);
17 // numArray.sumRange(0, 1);
18 // numArray.sumRange(1, 2);
posted @ 2015-11-13 09:32  Dylan_Java_NYC  阅读(314)  评论(0编辑  收藏  举报