【LeetCode-1109】航班预订统计＜Java版＞

//这里有 n 个航班，它们分别从 1 到 n 进行编号。 
//
// 我们这儿有一份航班预订表，表中第 i 条预订记录 bookings[i] = [i, j, k] 意味着我们在从 i 到 j 的每个航班上预订了 k 个座
//位。 
// 请你返回一个长度为 n 的数组 answer，按航班编号顺序返回每个航班上预订的座位数。
// 示例：
// 输入：bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
//输出：[10,55,45,25,25]
//
// 提示： 
// 1 <= bookings.length <= 20000
// 1 <= bookings[i][0] <= bookings[i][1] <= n <= 20000 
// 1 <= bookings[i][2] <= 10000 
// Related Topics 数组 数学
// ? 79 ? 0package com.asiainfo.leetcode.editor.cn;
//java:航班预订统计class P1109CorporateFlightBookings{
    public static void main(String[] args){
    Solution solution = new P1109CorporateFlightBookings().new Solution();int[][] ints = new int[3][];ints[0] = new int[]{
    1,2,10};ints[1] = new int[]{
    2,3,20};ints[2] = new int[]{
    2,5,25};solution.corpFlightBookings(ints,5);}//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
    int[] counters = new int[n];for(int[] booking : bookings){
    counters[booking[0]-1] += booking[2];if(booking[1] < n){
    }counters[booking[1]] -= booking[2];}for(int i = 0; i < counters.length; i++){
    counters[i+1] += counters[i];}return  counters;}
}
//leetcode submit region end(Prohibit modification and deletion)}