差分数组-leetcode1094

车上最初有 capacity 个空座位。车 只能 向一个方向行驶（也就是说，不允许掉头或改变方向）

给定整数 capacity 和一个数组 trips , trip[i] = [numPassengersi, fromi, toi] 表示第 i 次旅行有 numPassengersi 乘客，接他们和放他们的位置分别是 fromi 和 toi 。这些位置是从汽车的初始位置向东的公里数。

当且仅当你可以在所有给定的行程中接送所有乘客时，返回 true，否则请返回 false。

示例 1：

输入：trips = [[2,1,5],[3,3,7]], capacity = 4
输出：false

示例 2：

输入：trips = [[2,1,5],[3,3,7]], capacity = 5
输出：true

提示：

• 1 <= trips.length <= 1000
• trips[i].length == 3
• 1 <= numPassengersi <= 100
• 0 <= fromi < toi <= 1000
• 1 <= capacity <= 105

思路：差分数组，对于区间频繁修改的数组

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int n = 1000;
        int[] result = new int[n];
        for (int[] trip : trips) {
            int numPassengersi = trip[0];
            if (numPassengersi > capacity) {
                return false;
            }

            int start = trip[1] - 1;
            int end = trip[2] - 1 - 1;
            result[start] += numPassengersi;
            if (end + 1 < n) {
                result[end + 1] -= numPassengersi;
            }
        }

        for (int i = 0; i < 10; i++) {
            System.out.print(result[i] + "  ");
        }
        System.out.println();

        for (int i = 1; i < n; i++) {
            result[i] += result[i - 1];
        }

        for (int i = 0; i < 10; i++) {
            System.out.print(result[i] + "  ");
        }
        System.out.println();

        for (int i = 1; i < n; i++) {
            if (result[i] > capacity) {
                System.out.println(result[i] + "  " + capacity);
                return false;
            }
        }
        return true;


    }
}
//leetcode submit region end(Prohibit modification and deletion)