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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.
算法思路：
先找到哪些加油站可以安全起步，然后再逐个从这些安全的加油站出发。
代码如下：
1 public class Solution { 2 public int canCompleteCircuit(int[] gas, int[] cost) { 3 int[] stations = new int[gas.length]; 4 int count = gas.length,index = 0; 5 for(int i = 0; i < count;i++){ 6 if(gas[i] >= cost[i]){ 7 stations[index++] = i; 8 } 9 } 10 breakable: 11 for(int i = 0; i < index; i++){ 12 int gasCount = 0; 13 int costCount = 0; 14 for(int j = stations[i]; j < count + stations[i]; j++){ 15 gasCount += gas[j % count]; 16 costCount += cost[j % count]; 17 if(costCount > gasCount) continue breakable; 18 } 19 return stations[i]; 20 } 21 return -1; 22 } 23 }
[leetcode]Gas Station,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/huntfor/p/3891345.html
