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# 1109. Corporate Flight Bookings(M)

https\://leetcode.com/problems/corporate-flight-bookings/

There are `n` flights that are labeled from `1` to `n`.

You are given an array of flight bookings `bookings`, where `bookings[i] = [firsti, lasti, seatsi]` represents a booking for flights `firsti` through `lasti` (**inclusive**) with `seatsi` seats reserved for **each flight** in the range.

Return *an array* `answer` *of length* `n`*, where* `answer[i]` *is the total number of seats reserved for flight* `i`.

 

**Example 1:**

```
Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]
Explanation:
Flight labels:        1   2   3   4   5
Booking 1 reserved:  10  10
Booking 2 reserved:      20  20
Booking 3 reserved:      25  25  25  25
Total seats:         10  55  45  25  25
Hence, answer = [10,55,45,25,25]
```

**Example 2:**

```
Input: bookings = [[1,2,10],[2,2,15]], n = 2
Output: [10,25]
Explanation:
Flight labels:        1   2
Booking 1 reserved:  10  10
Booking 2 reserved:      15
Total seats:         10  25
Hence, answer = [10,25]
```

 

**Constraints:**

* `1 <= n <= 2 * 104`
* `1 <= bookings.length <= 2 * 104`
* `bookings[i].length == 3`
* `1 <= firsti <= lasti <= n`
* `1 <= seatsi <= 104`

### `Solution:`

这个题目就在那绕弯弯，其实它就是个差分数组的题，我给你翻译一下：

给你输入一个长度为 `n` 的数组 `nums`，其中所有元素都是 0。再给你输入一个 `bookings`，里面是若干三元组 `(i,j,k)`，每个三元组的含义就是要求你给 `nums` 数组的闭区间 `[i-1,j-1]` 中所有元素都加上 `k`。请你返回最后的 `nums` 数组是多少？

> PS：因为题目说的 `n` 是从 1 开始计数的，而数组索引从 0 开始，所以对于输入的三元组 `(i,j,k)`，数组区间应该对应 `[i-1,j-1]`。

这么一看，不就是一道标准的差分数组题嘛？我们可以直接复用刚才写的类：

```java
int[] corpFlightBookings(int[][] bookings, int n) {
    // nums 初始化为全 0
    int[] nums = new int[n];
    // 构造差分解法
    Difference df = new Difference(nums);

    for (int[] booking : bookings) {
        // 注意转成数组索引要减一哦
        int i = booking[0] - 1;
        int j = booking[1] - 1;
        int val = booking[2];
        // 对区间 nums[i..j] 增加 val
        df.increment(i, j, val);
    }
    // 返回最终的结果数组
    return df.result();
}
```

这道题就解决了。