- Car Pooling
Medium
There is a car with capacity empty seats. The vehicle only drives east (i.e., it cannot turn around and drive west).
You are given the integer capacity and an array trips where trips[i] = [numPassengersi, fromi, toi] indicates that the ith trip has numPassengersi passengers and the locations to pick them up and drop them off are fromi and toi respectively. The locations are given as the number of kilometers due east from the car’s initial location.
Return true if it is possible to pick up and drop off all passengers for all the given trips, or false otherwise.
Example 1:
Input: trips = [[2,1,5],[3,3,7]], capacity = 4
Output: false
Example 2:
Input: trips = [[2,1,5],[3,3,7]], capacity = 5
Output: true
Constraints:
1 <= trips.length <= 1000
trips[i].length == 3
1 <= numPassengersi <= 100
0 <= fromi < toi <= 1000
1 <= capacity <= 105
解法1：差分数组
注意：
- 这里i站上车，j站下车，实际上是在i…j-1之间累加x个乘客，不是i…j之间。
- 我们不知道具体多少个车站，所以根据input，直接定义diffs(1001), realNums(1001)。
- 在第一个for loop里面会记录每个车站的累积上车数，不会有遗漏的情况。
class Solution {
public:
bool carPooling(vector<vector<int>>& trips, int capacity) {
int n = trips.size();
vector<int> diffs(1001, 0);
int maxStation = 0;
for (int i = 0; i < n; i++) {
diffs[trips[i][1]] += trips[i][0];
maxStation = max(maxStation, trips[i][2]);
//if (trips[i][2] + 1 < n) {
diffs[trips[i][2]] -= trips[i][0];
//}
}
vector<int> realNums(1001, 0);
realNums[0] = diffs[0];
if (realNums[0] > capacity) return false;
for (int i = 1; i < maxStation; i++) {
realNums[i] = realNums[i - 1] + diffs[i];
if (realNums[i] > capacity) return false;
}
return true;
}
};