Leetcode-Gas Station

LiBlog發表於2014-12-23

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.

Analysis:

The amount of gas left in tank is non-decreasing. If we cannot pass a station P (>x) strating from x, it is because we do not accmulate enough gas in the tank, so we start from x+1, x+2,...P-1, we will still not be able to pass P. Instead, we try to start from P+1, so that we have the maximum accumulative gas in the tank when reaching P.


If we cannot pass P (<x) starting from x, we already try P+1 (<=x), then we have no way to pass P, fail directly.

Solution:

 1 public class Solution {
 2     public int canCompleteCircuit(int[] gas, int[] cost) {
 3         if (gas.length==0) return -1;
 4         if (gas.length==1)
 5             if (gas[0]-cost[0]>=0) return 0;
 6             else return -1;
 7 
 8         int len = gas.length;
 9         int left = 0;
10         int index = 0;
11         while (index<len){
12             left = 0;
13             //Check whether the current station is the solution.
14             boolean valid = true;
15             int p = index;
16             while (p!=(index+gas.length-1)%gas.length){
17                 if (left+gas[p]-cost[p]<0){
18                     valid = false;
19                     break;
20                 } else{
21                     left += gas[p]-cost[p];
22                     p = (p+1)%gas.length;
23                 }
24             }
25             if (valid && left+gas[p]-cost[p]>=0) return index;
26             else if ((p+1)%gas.length<=index) return -1;
27             else  index = p+1;
28         }
29 
30         return -1;
31     }
32 }