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.
首先，這個(gè)題目要明確如果gas[0] + gas[1] + ... + gas[n] >= cost[0] + cost[1] + .. + cost[n]，那么這個(gè)題目一定有解。因?yàn)椋鶕?jù)條件移項(xiàng)可得：
(gas[0] - cost[0]) + (gas[1] - cost[1]) + ... + (gas[n] - cost[n]) >= 0，由于最終結(jié)果大于等于零，因此一定可以通過加法交換律，得到一個(gè)序列，使得中間結(jié)果都為非負(fù)。因此可以將算法實(shí)現(xiàn)如下：
1 public class GasStation {
2 public int canCompleteCircuit(int[] gas, int[] cost) {
3 int sum = 0, total = 0, len = gas.length, index = -1;
4 for (int i = 0; i < len; i++) {
5 sum += gas[i] - cost[i];
6 total += gas[i] - cost[i];
7 if (sum < 0) {
8 index = i;
9 sum = 0;
10 }
11 }
12 return total >= 0 ? index + 1 : -1;
13 }
14 }