LeetCode 134. Gas Station 加油站（Java）

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 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.

解答：

解法一：

一开始采用了常规思维，即：

1. 遍历 i 个站点，假设从第 i 个站点开始出发，temp 表示从第 i 个站点出发走到的第 temp 个站点
2. 用 last 表示经过第 temp 个站点后油桶中剩余的油量。若剩余的油量 < 0，则说明无法到达第 temp + 1 个站点，则直接 break，接着遍历假设从第 i+1 个站点出发的情况
3. 若剩余的油量 > 0，则到达下一个 i+1 站点，更新剩余油量。
4. 其中，若到达最后一个站点，则回到第一个站点。若走完所有站点且 last>=0，则说明可以走完

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int len = gas.length;
        for(int i=0; i<len; i++) {  //假设从第0个站点开始出发，遍历
            int count =0;   //记录通过的站点数
            int last = gas[i];
            int temp = i;    //表示若从i开始出发，当前遍走到的第temp个站点
            while(count < len) {
                last -=  cost[temp];    //经过第temp个站点后，剩余的油量
                if(last < 0) {
                    break;
                }
                if(temp == len-1) {
                    last += gas[0];
                    temp = 0;
                } else {
                    last += gas[temp + 1];
                    temp++;
                }
                count++;
                if(count == len && last >= 0) {
                    return i;
                }
            }
        }
        return -1;
    }
}

最终提交结果为：

解法二：优化

之后看到了一种更优化的解法：他人的优化思路及解法
总结来说即：
若所有站点加的气 - 所有站点消耗的气 >= 0，则一定存在满足条件的站点。

所以我们在从头至尾遍历数组，寻找站点时，同时也可以获取到：所有站点加的气 - 所有站点消耗的气的值，从而在找到一个站点可以从 i 到达数组尾部后，不需要再返回头部继续检查，一定程度的降低了时间复杂度，从O(2n) 降低到了O(n).

代码实现中，用 total 表示所有站点加的气和消耗的气的总差值，用 sum 表示当前 i 出发时，油桶里的剩余值，便于找到出发点 i

代码实现如下：

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int start = 0;  //  出发位置
        int total = 0;  //所有gas值-所有cost值
        int sum = 0;    //从i出发情况下，当前桶中的剩余值
        for(int i=0; i<gas.length; i++) {
            total += gas[i] - cost[i];  //更新total值
            if(sum < 0) {
                sum = gas[i] - cost[i]; //重新赋值sum
                start = i;  //更新出发位置
            } else {
                sum += gas[i] - cost[i];
            }
        }
        return total >=0 ? start : -1;
    }
}

最终提交结果为：