Leetcode | Gas Station

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.

这道题也是做不出来。思路并没有错，只是中间有些证明并没有想清楚，所以一直卡住。google了一下，解题的详细思路如下：

1. If car starts at A and can not reach B. Any station between A and B can not reach B.(B is the first station that A can not reach.)
2. If the total number of gas is bigger than the total number of cost. There must be a solution.

证明： 1. 这里用left[i...j]来表示在从i到j累计剩下的油，也就是left[i...j]=(gas[i]-cost[i]) + ... _ (gas[j]-cost[j])。不能到达B，说明left[A...B]<0。也就是说，left[A...B-1]>0 but left[A...B-1] + (gas[B]-cost[B]) < 0，所以gas[B]-cost[B]<0，B也不是解。

注意从A能够到达B-1，所以left[A...i] > 0, A <= i < B。假设我们从一个新的位置C(A <= C <B)重新出发，已知 0< left[A...B-1]=left[A...C-1]+left[C...B-1]且left[A...C-1]>0，那么我们有left[C...B-1]<left[A...B-1]。因为left[A...B-1] + (gas[B]-cost[B]) < 0，所以left[C...B-1] + (gas[B]-cost[B]) < 0，也就是C也不能到达B。证毕。因此只能从B+1重新开始。

2. 反证法即可。假设不存在解，那么对于所有的0<=k<N，left[k...N-1,0...k-1]<0。但是由题设left[0...N-1]>0，矛盾。于是一定存在解。

于是，我们可以统计总的left，如果<0，无解。

否则，从每个位置i开始，看它能不能到达N-1，当它不能到达某个位置B时，也就是left[i...B]<0时，ii到B都不是解。我们从B+1继续搜索。最后一个cadidate一定就是解。

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int n = gas.size();
        int j = -1, sum = 0, total = 0;
        for (int i = 0; i < n; ++i) {
           sum += gas[i] - cost[i];
           total += gas[i] - cost[i];
           if (sum < 0) {
               j = i;
               sum = 0;
           }
        }
        if (total < 0) return -1;
        else return j + 1;
    }
};