Description
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.
Solution
DP, time O(n), space O(1)
Two things:
-
If car starts at A and can not reach B. Any station between A and B can not reach B.(B is the first station that A can not reach.) - If the total number of gas is bigger than the total number of cost. There must be a solution.
把gas[i] - cost[i]计算出来得到新的arr[]，然后感觉就变成了最经典的MaxSubarraySum的变种。累加的sum如果变成负值，就直接丢掉，以下一个起点开始重新累加。
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int start = 0;
int tank = 0;
int total = 0;
for (int i = 0; i < gas.length; ++i) {
if (tank < 0) {
tank = 0;
start = i;
}
tank += gas[i] - cost[i];
total += gas[i] - cost[i];
}
return total < 0 ? -1 : start;
}
}
