NumArray.java

package algorithms;
/**
 * 给定一个整数数组 nums，求出数组从索引i到j（i≤j）范围内元素的总和，包含i、j两点。
 * <p>
 * 实现 NumArray 类：
 * <p>
 * NumArray(int[] nums) 使用数组 nums 初始化对象
 * int sumRange(int i, int j) 返回数组 nums 从索引i到j（i≤j）范围内元素的总和，包含i、j两点（也就是 sum(nums[i], nums[i + 1], ... , nums[j])）
 * <p>
 * <p>
 * 示例：
 * <p>
 * 输入：
 * ["NumArray", "sumRange", "sumRange", "sumRange"]
 * [[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
 * 输出：
 * [null, 1, -1, -3]
 * <p>
 * 解释：
 * NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
 * numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
 * numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
 * numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
 */


/**
 * 解题思路
 * 转换 -> nums[i] = nums[0]+nums[1]+...+nums[i]
 * 则 nums[i,j] = nums[j]-nums[i] = nums[i+1]+...+nums[j]
 */
class NumArray {

    int[] nums;

    public NumArray(int[] nums) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            nums[i] = sum;
        }
        this.nums = nums;
    }

    public int sumRange(int left, int right) {
        if (left == 0) {
            return nums[right];
        } else {
            return nums[right] - nums[left - 1];
        }
    }

    public static void main(String[] args) {
        NumArray numArray = new NumArray(new int[]{-2, 0, 3, -5, 2, -1});
        System.out.println(numArray.sumRange(0, 2));// return 1 ((-2) + 0 + 3)
        System.out.println(numArray.sumRange(2, 5));// return -1 (3 + (-5) + 2 + (-1))
        System.out.println(numArray.sumRange(0, 5));// return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

    }
}