Gas Station(LeetCode)

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.

题目理解：N个加油站，形成一个环路，每个加油站i到i+1的耗油量为cost[i]，每个加油站可以补充的油量为gas[i]。求该环路中的一个起点，使得能够汽车能跑完全程。

举个例子：gas{1, 2, 4, 3, 0, 7} cost{4, 0, 2, 3, 5, 6}

从站i到达下一站能够贡献的剩余油量为 gas[i]-cost[i]， remain{-3, 2, 2, 0, –5, 7}，明显从剩余量为7的加油站开始，能够跑完整圈。

分析待续。。

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