#187 Gas Station

最新推荐文章于 2021-02-15 21:03:58 发布

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最新推荐文章于 2021-02-15 21:03:58 发布

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分类专栏： lintcode 文章标签： lintcode greedy

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本文介绍了一个使用贪婪算法解决的问题——寻找能够完成环路旅行的起始加油站。通过维护局部和全局油量差值，该算法能在O(n)时间内确定解决方案，并且仅使用O(1)额外空间。

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题目描述：

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 costscost[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.

Notice

The solution is guaranteed to be unique.

Have you met this question in a real interview?

Yes

Example

Given 4 gas stations with gas[i]=[1,1,3,1], and thecost[i]=[2,2,1,1]. The starting gas station's index is 2.

Challenge

O(n) time and O(1) extra space

题目思路：

这题用greedy算法，感觉不是那么正统的题。这题code中用了两个变量：local diff和global diff。local diff一直计算到i所费油量的总和，如果它<0，那么表示从前的那些gas stations都不是我们要找的，于是要把local diff和index reset，继续找后面的candidates。global diff为所有花费的油量总和，如果它<0，那么表示没有答案，return -1.

Mycode（AC = 37ms）：

class Solution {
public:
    /**
     * @param gas: a vector of integers
     * @param cost: a vector of integers
     * @return: an integer 
     */
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        // write your code here
        if (gas.size() == 0 || cost.size() == 0) {
            return -1;
        }
        
        int local_sumdiff = 0, global_sumdiff = 0;
        int idx = -1;
        
        for (int i = 0; i < gas.size(); i++) {
            local_sumdiff += gas[i] - cost[i];
            global_sumdiff += gas[i] - cost[i];
            
            // if local_sumdiff < 0, it means all the
            // previous gas stations are not candidates,
            // then reset the local_sumdiff and idx to
            // seek latter ones
            if (local_sumdiff < 0) {
                local_sumdiff = 0;
                idx = i;
            }
        }
        
        if (global_sumdiff < 0) {
            return -1;
        }
        else {
            return (idx + 1) % gas.size();
        }
        
    }
};