【題目】

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.

N個加油站一個圈，加油gas，耗費cost。

要求選一個加油站出發，可以開車經過所有加油站。

資料特點：具有唯一解，全正數，非空，等長。

【思路】

基於以下兩個很機智的結論

1、若第i個加油站，gas[i]-cost[i]<0，必不能從這裡出發。

2、因為加油站是一個圈，若ΣΔ(gas[i]-cost[i])<0無解。

【程式碼】

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int tank=0;//當前站出發後Δ裝油量
        int
 
 start=0;//出發站點
        int total=0;//迴圈總Δ油量
        for(int i=0;i<gas.length;i++){
            int delta=gas[i]-cost[i];
            tank+=delta;
            total+=delta;
            if(tank<0){
                tank=0;
                start=i+1;
                //站點i出發為負，必然只能從下一站點出發
            }
        }
        return total<0?-1:start;
    }
}

【舉例】

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.