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:

You may assume that the array does not change.

There are many calls to sumRange function.

1. class NumArray {
       private final int[] arr;
   
       public NumArray(int[] nums) {
           this.arr = new int[nums.length];
           int sum = 0;
           for(int i = 0; i < nums.length; i++){
               sum += nums[i];
               arr[i] = sum;
           }
       }
       
       public int sumRange(int i, int j) {
           return arr[j] - (i == 0 ? 0 : arr[i - 1]);
       }
   }

   arr[i]是nums数组从0到i（包括）的sum，所以要求i，j之间的和，先判断i是否为0，然后等于arr[j] - arr[i - 1]。

以上是dp解法。

倒是觉得easy题就要有easy题的亚子，直接设一个数组复制不就完了吗.

class NumArray {
    private final int[] arr;

    public NumArray(int[] nums) {
        arr = nums.clone();
    }
    
    public int sumRange(int i, int j) {
        int res = 0;
        for(int m = i; i <= j; i++){
            res += arr[i];
        }
        return res;
    }
}