leetcode--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:

You may assume that the array does not change.
There are many calls to sumRange function.

解：

想了半天怎么动态规划（打表）将sumRange在初始化的时候存到一个二元数组，怎么想都是O（ n2）n^2）

其实有可以O（n）初始化O（1）查询的解法

int[] nums;

public NumArray(int[] nums) {
for(int i = 1; i < nums.length; i++)
nums[i] += nums[i - 1];

this.nums = nums;
}

public int sumRange(int i, int j) {
if(i == 0)
return nums[j];

return nums[j] - nums[i - 1];
}