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.
解法:这道题的重点在于理解:当车从i跑不到j的时候,那么从i到j的中间任何一点也都跑不到j,因为能跑到某个站的前提是现有汽油数大于等于消耗数,因此这道题的思路就是,从0开始跑,直到跑不到i,那么就从i重新跑,直到从某个位置j能跑到0,这时候我们再检查从0能不能跑到j,如果可以就返回j,不可以则返回-1,代码如下:
public int canCompleteCircuit(int[] gas, int[] cost) {
int gass = gas[0], start = 0;
for (int i = 0; i < gas.length - 1; i++) {
gass -= cost[i];
if (gass >= 0) {
gass += gas[i+1];
} else {
start = i + 1;
gass = gas[i + 1];
}
}
gass -= cost[gas.length - 1];
if (gass < 0) {
return -1;
}
gass += gas[0];
for (int i = 0; i < start; i++) {
gass -= cost[i];
if (gass >= 0) {
gass += gas[i+1];
} else {
return -1;
}
}
return start;
}