问题描述:
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.
问题分析:
1.首先应该明确一点:若gas的总量>= cost的总量,那么一定存在一条解决路径。
2.寻找起点:设置起点res = 0;依次迭代到i,若res到i项cost和gas的和小于0,则证明res并不是真正的起点,重设res = i + 1,重复迭代过程。
过程详见代码:
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19class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int size = gas.size(); int sum = 0; int res = 0; int total = 0; for (int i = 0; i<size; ++i){ sum += gas[i] - cost[i]; if (sum<0){ total += sum; sum = 0; res = i + 1; } } total += sum; return total<0 ? -1 : res % size; } };
最后
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