Leetcode 303. Range Sum Query - Immutable

题目

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.

思路

最基础版的subarray sum query，直接使用prefix sum数组搞定。

复杂度

time O(n), space O(n)

代码

  public class NumArray {
    HashMap<Integer, Integer> map;
    public NumArray(int[] nums) {
        //edge case
        if (nums == null || nums.length == 0) {
            return;
        }
        map = new HashMap<Integer, Integer>();
        int prefixSum = 0;
        for (int i = 0; i < nums.length; i++) {
            prefixSum += nums[i];
            map.put(i, prefixSum);
        }
    }

    public int sumRange(int i, int j) {
        int shorterSum = i == 0? 0: map.get(i - 1);
        int longerSum = map.get(j);
        return longerSum - shorterSum;
    }
}


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