[LeetCode303] 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:

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

solution:

can use list to pre-process the input array, which store the left element sum. So once query range sum(i,j), left(j) - left(i-1), take O(1) to get result.


public class NumArray {
    ArrayList<Long> leftSum = new ArrayList<>();
    public NumArray(int[] nums) {
        long sum = 0;
        for(int i=0;i<nums.length;i++) {
            sum+= nums[i];
            leftSum.add(sum);
        }
    }

    public int sumRange(int i, int j) {
        if(i>0) {
            return (int)(leftSum.get(j) - leftSum.get(i-1));
        }else if(i==0) {
            return leftSum.get(j).intValue();
        }else {
            return 0;
        }
        
    }
}


// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);




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