leetcode || 134、Gas Station

problem：

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.

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Greedy

题意：汽车从某一个加油站出发，加油、消耗，找出唯一那个可以回到原点的加油站

thinking：

（1）算法比较简单，汽油的余量为负数时行不通，贪心策略

（2）我刚开始采用的DFS+剪分支的方式，提交超时，但这个思路适用面更广

（3）采用数组处理的方式可以避免函数递归调用的开销

code：

DFS+剪分支：超时

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int dep=0;
        int maxDep=gas.size()-1;
        if(maxDep<0)
            return 0;
        int fuel=0;
        bool flag=true;
        for(int index=0;index<gas.size();index++)
        {
            dfs(dep,maxDep,index,fuel,gas,cost,flag);
            if(flag)
            return index;
            flag=true;
        }
        return -1;
    }
protected:
    void dfs(int dep,int maxDep,int index,int fuel, vector<int> &gas, vector<int> &cost, bool &flag)
    {
        if(!flag)
            return;
        fuel+=gas[index];
        fuel-=cost[index];
        if(fuel<0)
        {
            flag=false;
            return;
        }
        if(dep==maxDep)
        {
            if(fuel<0)
                flag=false;
            else
                flag=true;
            return;
        }
        dfs(dep+1,maxDep,(index+1)%gas.size(),fuel,gas,cost,flag);
    }

};

贪心策略：AC

时间复杂度O（n）

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