LeetCode :: Gas Station
2014-07-19 22:50
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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.
通过gas - cost 构造一个数组left,left[i]表征 i 站到达下一站之后的剩余gas量;
注意点:1.记得把pos取模,从超出位置拉回来就好; 2. 根据聚合分析,时间复杂度是O(N),因为在一遍扫描之后就可以确定start位置,之后再追加一个N次扫描,因此至多为2N的时间。代码如下:
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.
通过gas - cost 构造一个数组left,left[i]表征 i 站到达下一站之后的剩余gas量;
注意点:1.记得把pos取模,从超出位置拉回来就好; 2. 根据聚合分析,时间复杂度是O(N),因为在一遍扫描之后就可以确定start位置,之后再追加一个N次扫描,因此至多为2N的时间。代码如下:
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int m = gas.size(); int count = 0, sum = 0, pos = 0, start = 0; vector<int> left; for (int i = 0; i < m; i++) left.push_back(gas[i] - cost[i]); for(int i = 0; i < m; i++) sum += left[i]; if (sum < 0) return -1; sum = 0; while (count < m){ //利用count来记录走过的站点数 sum += left[pos]; pos = ++pos % m; //取模来把超出范围的pos值拉回到0, m % m = 0 既第m+1 个pos又置0 if (sum < 0){ sum = 0; count = 0; start = pos; } else{ count++; } } return start; } };
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