leetcode_Gas Station
2015-05-16 08:21
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描述:
There are N gas stations along a circular route, where the amount of gas at station i is
You have a car with an unlimited gas tank and it costs
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.
思路:
这是一个动态规划的题目,假设能跑完全程,最后邮箱中的油肯定还有剩余,由于是一个环形的路线,所以理论来讲,可以从任何地方开始。所以,可以用两个变量来表示这道题目,sum表示全程的油的剩余,而index来表示从何处开始。
1.从任何地方可以开始本次的行程,而sum+=gas[i]-cost[i],若sum<0,说明从index开始是不合适的,换成从i+1开始,直至结尾
2.若最后sum>=0,说明路线存在且以index为起点的路线为一个可行的路线。
代码:
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.
思路:
这是一个动态规划的题目,假设能跑完全程,最后邮箱中的油肯定还有剩余,由于是一个环形的路线,所以理论来讲,可以从任何地方开始。所以,可以用两个变量来表示这道题目,sum表示全程的油的剩余,而index来表示从何处开始。
1.从任何地方可以开始本次的行程,而sum+=gas[i]-cost[i],若sum<0,说明从index开始是不合适的,换成从i+1开始,直至结尾
2.若最后sum>=0,说明路线存在且以index为起点的路线为一个可行的路线。
代码:
public int canCompleteCircuit(int[] gas, int[] cost) { int sum=0,tempSum=0,num=0,stationNum=0; int len=gas.length; for(int i=0;i<len;i++) { num=gas[i]-cost[i]; sum+=num; tempSum+=num; if(tempSum<0) { stationNum=(i+1)%len; tempSum=0; } } if(sum>=0) return stationNum; return -1; }
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