【LeetCode】Gas Station

最新推荐文章于 2024-10-12 23:30:39 发布

原创最新推荐文章于 2024-10-12 23:30:39 发布 · 481 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

86 篇文章

订阅专栏

动态规划

17 篇文章

订阅专栏

本文提供了一种使用C++解决环形油站问题的方法，包括核心内容解析、代码实现及关键信息概述。通过分析，确定从哪一站开始旅行才能完成环形路线，且总油量始终足够。同时，推荐了C++标准函数库、API查询及vector使用方法的学习资源。

参考链接

http://blog.youkuaiyun.com/jellyyin/article/details/12245429

http://blog.youkuaiyun.com/doc_sgl/article/details/12980297

题目描述

Gas Station

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.

题目分析

1. 每一站的代价为gas-cost, 也就是求从哪一站开始累加代价和总是大于0。

2. 如果所有站的代价和大于0，则所求的路线必定存在。

如果总代价〉=0,从序号0开始求代价和，如果代价和小于0，则不是从本站或者本站之前的某一个代价大于0的站开始，必从下一站即之后的站开始，而且这样的站必定存在。

总结

代码示例

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        // Note: The Solution object is instantiated only once.
		int total = 0;
		int currentgas = 0; 
		int startpoint = -1;
		int sz = gas.size();
		for(int i = 0; i < sz; i++)
		{
			currentgas += gas[i] - cost[i];
			total += gas[i] - cost[i];
			if(currentgas < 0)
			{
				startpoint = i;
				currentgas = 0;
			}
		}
		return total >= 0 ? startpoint+1 : -1;
    }
};