title: 1094. Car Pooling
tags:
- Prefix Sum
categories: leetcode
comments: false
class Prefix{
private:
vector<int> diff;
public:
Prefix(vector<int>& nums){
int n = nums.size();
diff = vector<int>(n, 0);
diff[0] = nums[0];
for(int i=1;i<n;++i){
diff[i] = nums[i] - nums[i-1];
}
}
void increment(int i, int j, int val){
diff[i]+=val;
if(j+1<diff.size()) diff[j+1]-=val;
}
vector<int> reconstruct(){
// 還原原本陣列
//diff 0 2 0 3 0 0 -2 0 -3 0
//ret 0 2 2 5 5 5 3 3 0 0
vector<int> ret(1001,0);
ret[0] = diff[0];
for(int i=1;i<1001;++i){
ret[i] = ret[i-1] + diff[i];
}
return ret;
}
};
class Solution {
public:
bool carPooling(vector<vector<int>>& trips, int capacity) {
//diff 0 0 0 0 0 0 0 0 0 0
//diff 0 2 0 0 0 -2 0 0 0 0
//diff 0 2 0 3 0 -2 0 -3 0 0
vector<int> diff(1001,0);
Prefix prefix = Prefix(diff);
for(vector<int>& trip:trips){
int i = trip[1], j = trip[2]-1, val = trip[0];
// 即乘客車上的區間是 [trip[1], trip[2] -1]
prefix.increment(i,j,val);
}
vector<int> ret = prefix.reconstruct();
// 客車自始至终都不應該超载
for(int r:ret){
if(r>capacity) return false;
}
return true;
}
};
- time complexity
O(n)
, n is number of trip ,O(n*1000)
- space complexity
O(n)