算法——Week15

134. 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 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.

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解题思路
满足条件的起始点，一定会有gas[i] > cost[i]，找出所有满足这一条件的点（最大的差值点不一定可以），依次计算是否可以走完一圈。

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代码如下：

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        vector<int> differ(gas.size());
        int max = 0;
        for(int i = 0; i < gas.size(); i++) {
            differ[i] = gas[i] - cost[i];
            if(differ[i] >= max) {
                max = differ[i];
            }
            cout << differ[i] << " ";
        }
        cout <<endl;
        vector<int> start;
        for(int i = 0; i < gas.size(); i++) {
            if(differ[i] >= 0) {
                start.push_back(i);
            }
        }

        for(int i = 0; i <start.size(); i++) {
            int sum = differ[start[i]];
            for(int t = 0; t < differ.size(); t++) {
                sum += differ[(start[i] + t + 1)%(differ.size())];
                //cout << (start[i] + t + 1)%(differ.size()) <<" ";
                if(sum < 0) {
                    break;
                }
            }
            if(sum >= 0)
                return start[i];
        }
        return -1;
    }
};

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上面的想法很简单但是很费时。对原来的算法进行改进，考虑到，如果A开始到B时sum < 0 ，那么AB之间所有的节点都不行，从B后面的那个节点开始新的遍历。

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