LintCode 187: Gas Station (Greedy算法经典题)

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博客围绕加油站问题展开，给定环形路线上各加油站的油量和行驶到下一站的耗油量，需判断能否绕一圈并找出起始加油站索引。介绍了使用贪心算法解决该问题，指出判断能否跑一圈的条件是totalSum > 0，还给出了示例及复杂度要求。

Gas Station
中文English
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.

Example
Example 1:

Input:gas[i]=[1,1,3,1],cost[i]=[2,2,1,1]
Output:2
Example 2:

Input:gas[i]=[1,1,3,1],cost[i]=[2,2,10,1]
Output:-1
Challenge
O(n) time and O(1) extra space

Notice
The solution is guaranteed to be unique.

解法1：Greedy Algorithm.
注意：
1）判断能不能跑一圈的条件就是totalSum > 0!
代码如下：

class Solution {
public:
    /**
     * @param gas: An array of integers
     * @param cost: An array of integers
     * @return: An integer
     */
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int N = gas.size();

        if (N == 0) return -1;
        int gasSum = 0;
        int candidate = 0;
        int totalSum = 0;

        for (int i = 0; i < N; ++i) {
            gasSum += gas[i];

            if (gasSum < cost[i]) {
                gasSum = 0;
                candidate = (i == N) ? 0 : i + 1;
            } else {
                gasSum -= cost[i];
            }

            totalSum += gas[i] - cost[i];
        }
        candidate = candidate % N;
        if (totalSum < 0) return -1;
        return candidate;
    }
};