303. Range Sum Query

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

class NumArray {
    
    int[] sums;
    public NumArray(int[] nums) {
        sums = new int[nums.length];
        if (nums.length == 0) return;
        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];  
    }
}