[Leetcode]-gas Station

There was N gas stations along a circular route, where the amount of gas at station I was gas[i].

You had a car with an unlimited gas tank and it costs cost[i] of the gas-to-travel from station I to its next station (I+1). You begin the journey with a 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.
Hide Tags:greedy
Title: Cycle a circle of the road there are gassize gas station gas[gassize], from I to i+1 stations need to consume cost[i] gas, requirements can be from any of the gas station, as long as can guarantee back to the starting point.
Idea: the greedy algorithm.
The first cycle: from gas[0] Add gas station I from 0-> gasSize-1, stand to judge whether can go to station I, over for each station after the remaining gas, over a negative description of the current station I, when found not to the current station, immediately will be the current station I as the starting point , over zeroed, and started to move forward again.
The second cycle: Of course, it is possible to walk after a circle can not be judged, because it is possible that the starting point is not gas[0], it is necessary to continue to move forward, until Gas[start], the period as long as there is no case this judgment is-1. If the last to determine whether over is greater than 0, greater than 0 will be able to, otherwise return-1.

int Cancompletecircuit (int* gas, int. gassize, int* cost, int costsize) {int i =0; Int Over= gas[0]; int start =0;if(Gassize = =1)return(gas[0] >= cost[0]) ?0: -1; for(i=1; i<gassize;i++) { Over= Over-cost[i-1];if( Over<0) {start = i; Over=0; } Over= Over+ Gas[i]; } Over= Over-cost[gassize-1]; for(i=0; i<start;i++) {if( Over<0)        {return-1; } Over= Over+ gas[i]-cost[i]; }return( Over>=0) ? Start:-1; }

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[Leetcode]-gas Station