暴力解法：按照示例在firsti到lasti遍历，将seati加到每个数上。时间复杂度取决于bookings的长度

一维差分：建立初始数组和差分数组，在差分数组的firsti位置加上seati，差分数组的lasti+1位置减去seati（避免影响原数组lasti之后的数），最后还原数组

代码

import org.junit.Test;

public class CountFights {

    @Test
    public void test() {

        int[][] bookings = new int[][]{{1, 2, 10}, {2, 3, 20}, {2, 5, 25}};
        int[] result = new int[5];
        result = countFights(bookings, 5);
        for (int i : result) {
            System.out.print(i + " ");//[10,55,45,25,25]
        }
        System.out.println();
        int[][] bookings1 = new int[][]{{1, 2, 10}, {2, 2, 15}};
        int[] result1 = new int[2];
        result1 = countFights(bookings1, 2);
        for (int i : result1) {
            System.out.print(i + " ");//[10,25]
        }


    }

    public static int[] countFights(int[][] bookings, int n) {
        int[] result = new int[n];//初始数组
        int[] difference = new int[n + 1];//差分数组，差分数组长度加一避免最后一位对后一位操作时越界
        for (int i = 0; i < bookings.length; i++) {
            //firsti和endi与差分数组的索引相差一
            difference[bookings[i][0] - 1] += bookings[i][2];//difference[firsti-1] += seati;
            difference[bookings[i][1]] -= bookings[i][2];//difference[endi] -= seati;
        }

        result[0] = difference[0];
        for (int i = 1; i < n; i++) {
            result[i] = result[i - 1] + difference[i];//还原数组
        }
        return result;
    }


}