差分?jǐn)?shù)組-leetcode1094
2023-04-12
車(chē)上最初有capacity個(gè)空座位。車(chē)只能向一個(gè)方向行駛(也就是說(shuō),不允許掉頭或改變方向)
給定整數(shù)capacity和一個(gè)數(shù)組trips,trip[i] = [numPassengersi, fromi, toi]表示第i次旅行有numPassengersi乘客,接他們和放他們的位置分別是fromi和toi。這些位置是從汽車(chē)的初始位置向東的公里數(shù)。
當(dāng)且僅當(dāng)你可以在所有給定的行程中接送所有乘客時(shí),返回true,否則請(qǐng)返回false。
示例 1:
輸入:trips = [[2,1,5],[3,3,7]], capacity = 4
輸出:false
示例 2:
輸入:trips = [[2,1,5],[3,3,7]], capacity = 5
輸出:true
提示:
1 <= trips.length <= 1000trips[i].length == 31 <= numPassengersi <= 1000 <= fromi < toi <= 10001 <= capacity <= 105
思路:差分?jǐn)?shù)組,對(duì)于區(qū)間頻繁修改的數(shù)組
//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)
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