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
差分数组。
class Solution:
def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
delta = [0] * 1001
for num, start, end in trips:
delta[start] += num
delta[end] -= num
return all(s <= capacity for s in accumulate(delta))
class Solution {
public boolean carPooling(int[][] trips, int capacity) {
int[] delta = new int[1001];
for (int[] trip : trips) {
int num = trip[0], start = trip[1], end = trip[2];
delta[start] += num;
delta[end] -= num;
}
int cur = 0;
for (int num : delta) {
cur += num;
if (cur > capacity) {
return false;
}
}
return true;
}
}
/**
* @param {number[][]} trips
* @param {number} capacity
* @return {boolean}
*/
var carPooling = function (trips, capacity) {
let delta = new Array(1001).fill(0);
for (let [num, start, end] of trips) {
delta[start] += num;
delta[end] -= num;
}
let s = 0;
for (let num of delta) {
s += num;
if (s > capacity) {
return false;
}
}
return true;
};
class Solution {
public:
bool carPooling(vector<vector<int>>& trips, int capacity) {
vector<int> delta(1001);
for (auto& trip : trips) {
int num = trip[0], start = trip[1], end = trip[2];
delta[start] += num;
delta[end] -= num;
}
int cur = 0;
for (auto& num : delta) {
cur += num;
if (cur > capacity) {
return false;
}
}
return true;
}
};
func carPooling(trips [][]int, capacity int) bool {
delta := make([]int, 1010)
for _, trip := range trips {
num, start, end := trip[0], trip[1], trip[2]
delta[start] += num
delta[end] -= num
}
cur := 0
for _, num := range delta {
cur += num
if cur > capacity {
return false
}
}
return true
}