[LeetCode]Gas Station

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题目描述：(链接)

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.

解题思路：

如果所有花费的汽油之和>所有站点可提供的汽油之和，无解， return -1，这里用gap变量记录该差值，并且用sum变量记录当前差值，如果差值小于0，将当前站点设为index，sum设为0。

 1 class Solution {
 2 public:
 3     int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
 4         int gap = 0;
 5         int sum = 0;
 6         int index = -1;
 7         for (int i = 0; i < gas.size(); ++i) {
 8             gap += gas[i] - cost[i];
 9             sum += gas[i] - cost[i];
10             if (sum < 0) {
11                 index = i;
12                 sum = 0;
13             }
14         }
15         
16         if (gap >= 0) {
17             return index + 1;
18         }
19         
20         return -1;
21     }
22 };

[LeetCode]Gas Station

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原文地址：http://www.cnblogs.com/skycore/p/4888508.html