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.

解题思路

考虑到要多次调用sumRange()函数，因此需要把结果先存起来，调用时就可以直接返回了。最开始考虑的是用dp[i][j]来直接存储i到j之间元素的和，但是内存超出限制。于是考虑用dp[i]来存储0到i之间元素的和，0到j的和减去0到i-1的和即为所求。

实现代码 NumArray { private int[] dp; public NumArray(int[] nums) { dp = new int[nums.length]; int sum = 0; for (int i = 0; i < nums.length; i++) { sum += nums[i]; dp[i] = sum; } } (int i, int j) { return i == 0 ? dp[j] : dp[j] - dp[i - 1]; } }

作者：u011331383 发表于2015/11/10 15:09:10 原文链接

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