題目如下：
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.hild.
我的提交：
/*17.3.31
*有點像最大連續(xù)子序列和的動態(tài)規(guī)劃解法
*如果total大于0影斑，從左到右掃描，一定能切分出一個和大于0的區(qū)間（起點為i,終點為數(shù)組的結(jié)尾）
*如-1控淡，-2慎式，2赛惩，3，-6，10
*-1被切開散休，-2被切開次屠，3园匹，4，-6被切開劫灶，10大于0裸违，所以以10為起點！
*換一個順序-6浑此，10累颂，-1，-2凛俱，2紊馏，3
*-6被切開，10~3大于0蒲犬，結(jié)果還是以10為起點朱监。
*/
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int start = 0,total = 0;
for(int i=0,tank=0;i<gas.length;i++){
if(tank<0){
tank = gas[i] - cost[i];
start = i;
}
else{
tank += gas[i] - cost[i];
}
total += gas[i] - cost[i];
}
return total<0?-1:start;
}
}
