LeetCode(303) Range Sum Query - Immutable解题报告

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:

直接求解会超时,所以采用离线算法，先将结果保存到数组中，用的过程中直接调用即可，这里有几点注意的情况：

1.输入数组为空

2.查询从零或者不从零开始

3.保存Sums数组是用的sums[i] = sums[i-1] + nums[i] (一开始没注意，通过循环存和，结果超时！囧)

附上代码:

public class NumArray {
private int[] sums;
public NumArray(int[] nums) {
if(nums.length == 0)
return;
sums = new int[nums.length];
sums[0] = nums[0];
for(int i = 1; i < nums.length; i++){
sums[i] = sums[i-1] + nums[i];
}
}

public int sumRange(int i, int j) {
return i == 0 ? sums[j] : sums[j] - sums[i-1];
}