Leetcode:Gas Station
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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. 假设出发位置start初始为gas.size()-1;可到达位置(油箱内有足够)end初始为0。
2. 车内油量=车站油量-消耗,即从出发位置开始:sum = gas[start] - cost[start];
3. 如果sum>=0,则表示可以到达下一车站,车内油量:sum += gas[end] - cost[end];
此时把可到达位置前进一格,执行end++操作;
如果sum<0,则表示从该位置出发不能够行使一周,此时,出发位置退后一格,执行start--操作;
车内油量:sum += gas[start] - cost[start];
4. 如果start>end,重复步骤3。
5. 最后,如果sum>=0,返回start,否则返回-1。
实现代码:
class Solution {public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int start = gas.size()-1; int end = 0; int sum = gas[start] - cost[start]; while (start > end) { if (sum >= 0) { sum += gas[end] - cost[end]; ++end; } else { --start; sum += gas[start] - cost[start]; } } return sum >= 0 ? start : -1; }};
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