leetcode Gas Station
Gas Station
题目:
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.
解析:
1 计算空车从某个station出发,到达下一个station的gasLeft。
2 以某个station为起点进行遍历。
3 计算从起点到达每个station的gasLeft,如果有负值,则不符合条件。
4 想不到竟然Accept了。
代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 private boolean canCompleteCircuit(int[] gasLeft, int startStation) { 2 int gas = 0; 3 int currentStation = startStation; 4 while(true) { 5 gas += gasLeft[currentStation]; 6 if (gas < 0) { 7 return false; 8 } 9 currentStation ++; 10 if (currentStation == gasLeft.length) { currentStation = 0; } 11 if (currentStation == startStation) { break; } 12 } 13 return true; 14 } 15 public int canCompleteCircuit(int[] gas, int[] cost) { 16 int[] gasLeft = new int[gas.length]; 17 for (int i = 0; i < cost.length; i++) { 18 gasLeft[i] = gas[i] - cost[i]; 19 } 20 for (int i = 0; i < gasLeft.length; i++) { 21 if (gasLeft[i] >= 0 && canCompleteCircuit(gasLeft, i)) { 22 return i; 23 } 24 } 25 return -1; 26 }
原本以为会超时。
其实最开始想到的是,汽车可以选择方向。先判断可能travel around的方向,再进行遍历。
这题看起来会有更好解法,请留言。
全文完。
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重剑无锋,大巧不工。