134. Gas Station (Array; DP)

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.

思路：

1. 是否能travel around? =>只要gas[]加总的和 - cost[]加总的和 > 0，就能够。=>需要一个INT计算gas[]-cost[]的总合

2. 如何找到起点？如果到了某一站油不够，那么起点必须从它后面开始重新找。=>需要一个INT记录从找到的起点到目前为止的油量

时间复杂度O(n)

class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int sum = 0, total = 0, len = gas.size(), index = -1;
for(int i=0; i<len; i++){
sum += gas[i]-cost[i];
total += gas[i]-cost[i];
if(sum < 0){
index = i;
sum = 0;
}
}
return total>=0 ? index+1 : -1;
}
};