LeetCode -- Gas Station
題目描述:
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.
Note:
The solution is guaranteed to be unique.
即有n個gas站,到達第i個站的消耗為cost[i],能否找到一個出發點,可以跑完gas[0...n)一圈。
解法一:
分別以每個加油站作為出發點(stopNo)進行測試 , StopNo ∈gas[i,n):
delta= gasInCar(初始化為0) + gas[i] - cost[i]
如果delta >= 0,累加gasInCar,去下一站i,累加成功到達下一站的次數:counter。如果counter達到gas.Length,返回i。
否則,重置gasInCar = 0,stopNo++。i=下一站,counter重置為0。
由於解法為O(N^2)的時間復雜度,無法通過測試數據:
代碼:
public int CanCompleteCircuit(int[] gas, int[] cost) {
var gasInCar = 0;
var stopNo = 0;
var i = 0;
var counter = 0;
while(stopNo < gas.Length){
if(gas[i] + gasInCar < cost[i]){
// failed , start at a new place and re count again
gasInCar = 0;
stopNo ++;
i = NextStop(stopNo, gas.Length);
counter = 0;
}
else{
gasInCar += gas[i]-cost[i];
i = NextStop(i, gas.Length);
counter ++;
}
if(counter == gas.Length){
return i;
}
}
return -1;
}
private int NextStop(int stop,int len){
if(stop < len - 1){
stop ++;
}
else {
stop = 0;
}
return stop;
}
解法二:
將此題分解為兩個小問題:
1.是否可完成一圈
2.找到出發點
存車中剩余gas為gasInCar=0,delta為當前站到下一站的gas[i]-cost[i],startAt為出發點。
一次遍歷gas[i...n-1] :
如果delta >= 0 : gasInCar += delta(積累車中gas)
如果小於0(即不可達):gasInCar 設為 delta,startAt = i,即重置當前點為出發點以及車中的gas
每次積累totalDelta
最後判斷totalDelta 是否大於0=能否完成一圈。
實現代碼:
public int CanCompleteCircuit(int[] gas, int[] cost)
{
var gasInCar = 0;
var totalDelta = 0;
var startAt = 0;
for (var i = 0; i < gas.Length; i++) {
var delta = gas[i] - cost[i];
if (gasInCar >= 0) {
gasInCar += delta;
} else {
gasInCar = delta;
startAt = i;
}
totalDelta += delta;
}
if (totalDelta >= 0){
return startAt;
}else{
return -1;
}
}