LeetCode_303. Range Sum Query - Immutable

303. Range Sum Query - Immutable

Easy

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.

package leetcode.easy;

class NumArray1 {
	private int[] data;

	public NumArray1(int[] nums) {
		data = nums;
	}

	public int sumRange(int i, int j) {
		int sum = 0;
		for (int k = i; k <= j; k++) {
			sum += data[k];
		}
		return sum;
	}
}

class NumArray3 {
	private int[] sum;

	public NumArray3(int[] nums) {
		sum = new int[nums.length + 1];
		for (int i = 0; i < nums.length; i++) {
			sum[i + 1] = sum[i] + nums[i];
		}
	}

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

/**
 * Your NumArray object will be instantiated and called as such: NumArray obj =
 * new NumArray(nums); int param_1 = obj.sumRange(i,j);
 */
public class RangeSumQueryImmutable {
	@org.junit.Test
	public void test1() {
		int[] nums = { -2, 0, 3, -5, 2, -1 };
		NumArray1 obj = new NumArray1(nums);
		int param_1 = obj.sumRange(0, 2);
		int param_2 = obj.sumRange(2, 5);
		int param_3 = obj.sumRange(0, 5);
		System.out.println(param_1);
		System.out.println(param_2);
		System.out.println(param_3);
	}

	@org.junit.Test
	public void test3() {
		int[] nums = { -2, 0, 3, -5, 2, -1 };
		NumArray3 obj = new NumArray3(nums);
		int param_1 = obj.sumRange(0, 2);
		int param_2 = obj.sumRange(2, 5);
		int param_3 = obj.sumRange(0, 5);
		System.out.println(param_1);
		System.out.println(param_2);
		System.out.println(param_3);
	}
}

原文地址：https://www.cnblogs.com/denggelin/p/11786009.html