[leetcode] GAS station
2014-04-18 10:26
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There are N gas stations along a circular route, where the amount of gas at station i is
You have a car with an unlimited gas tank and it costs
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. 每一站的代价为gas-cost, 也就是求从哪一站开始累加代价和总是大于0。一开始用了一个O(n^2)的解法,超时
2. 如果所有站的代价和大于0,则所求的路线必定存在。如果总代价〉=0,从序号0开始求代价和,如果代价和小于0,则不是从本站或者本站之前的某一个代价大于0的站开始,必从下一站即之后的站开始,而且这样的站必定存在O(n)
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. 每一站的代价为gas-cost, 也就是求从哪一站开始累加代价和总是大于0。一开始用了一个O(n^2)的解法,超时
2. 如果所有站的代价和大于0,则所求的路线必定存在。如果总代价〉=0,从序号0开始求代价和,如果代价和小于0,则不是从本站或者本站之前的某一个代价大于0的站开始,必从下一站即之后的站开始,而且这样的站必定存在O(n)
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { // Note: The Solution object is instantiated only once and is reused by each test case. int sum=0; int total=0; int start=0; for(int i=0;i<gas.size();i++) { sum+=gas[i]-cost[i]; total+=gas[i]-cost[i]; if(sum < 0) { start=(i+1)%gas.size(); sum=0; } } if(total <0) return -1; else return start; } };
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