问题描述
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 in the clockwise direction, otherwise return -1.
Note:
If there exists a solution, it is guaranteed to be unique. Both input arrays are non-empty and have the same length. Each element in the input arrays is a non-negative integer. Example 1:
Input: gas = [1,2,3,4,5] cost = [3,4,5,1,2]
Output: 3
Explanation: Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 - 1 + 5 = 8 Travel to station 0. Your tank = 8 - 2 + 1 = 7 Travel to station 1. Your tank = 7 - 3 + 2 = 6 Travel to station 2. Your tank = 6 - 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index. Example 2:
Input: gas = [2,3,4] cost = [3,4,3]
Output: -1
Explanation: You can’t start at station 0 or 1, as there is not enough gas to travel to the next station. Let’s start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 0. Your tank = 4 - 3 + 2 = 3 Travel to station 1. Your tank = 3 - 3 + 3 = 3 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. Therefore, you can’t travel around the circuit once no matter where you start.
解决思路
从第一个加油站开始走，用left记录剩余油量，如果left一直大于0，没问题；如果left小于0，则从下一个位置重新计算。并且，用lack记录之前所有欠下的油量，如果最后left - lack大于0，则说明可以有满足要求的出发点；若left - lack小于0，则说明没有满足要求的，就返回-1.
问题解惑
- 为什么
left < 0，不是从之前的某一个点计算，而是从当前点的下一个开始计算？
前提：若从0出发，走到第n个点时，left<0。
假设：中间某个点m可以作为出发点。
因为m可以作为出发点，则说明到m点时，left>0。所以若满足走到n时，left<0的前提，则从m到n这个过程中总的(cost[i]-gas[i])<0,则说明从m不可能到达n，这与前提“从0出发，可以走到第n个点”矛盾。 结论：当前点之前的所有点都不满足要求。
代码
package leetcode;
public class GasStation {
public int canCompleteCircuit(int[] gas, int[] cost) {
int left = 0, lack = 0, start = 0;
for (int i = 0; i < gas.length; i++) {
left += gas[i] - cost[i];
if(left < 0){
start = i + 1;
left = 0;
lack += left;
}
}
return left - lack >= 0 ? start: -1;
}
}
