[LC] 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:

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

Solution 1:

class NumArray {

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

    public int sumRange(int i, int j) {
        if (i == 0) {
            return prefix[j];
        }
        return prefix[j] - prefix[i - 1];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

Solution 2:

class NumArray {
    int[] prefix;
    public NumArray(int[] nums) {
        prefix = new int[nums.length + 1];
        for (int i = 0; i < nums.length; i++) {
            prefix[i + 1] = prefix[i] + nums[i];
        }
    }

    public int sumRange(int i, int j) {
        return prefix[j + 1] - prefix[i];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

原文地址：https://www.cnblogs.com/xuanlu/p/12008635.html