Gas Station@leetcode

刷题必备书籍：Cracking the Coding Interview: 150 Programming Questions and Solutions 

简历：The Google Resume: How to Prepare for a Career and Land a Job at Apple, Microsoft, Google, or any Top Tech Company
算法学习书籍：Introduction to Algorithms
编程珠玑：Programming Pearls (2nd Edition)
C++ 学习：The C++ Programming Language, 4th Edition
经典操作系统书籍，龙书：Operating System Concepts
创业：The Start-up of You: Adapt to the Future, Invest in Yourself, and Transform Your Career

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.

题没什么难的，就是要注意这是一个circular route, i后面连接的是0，所以index需要处理，而走了多远要用一个额外的变量记录。

刚开始一直time limit exceed，因为我是一个一个station作为start去尝试，后来想其实没必要，因为假如一旦有一个route走不通，我们重新开始的station应该是我们中断的下一个，因为前面的任何一个开始都不会可以，可以证明出来，不细说了，自己想想就可以知道。

	class Solution {
	public:
	int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
	// Note: The Solution object is instantiated only once and is reused by each test case.
	int gasVal=0;
	int len=gas.size();
	for(int i=0; i<gas.size(); i++)
	{
	int j=i;
	int count = 1;
	gasVal=0;
	while(count<=gas.size())
	{
	if(gasVal+gas[j%len]>=cost[j%len])
	{
	if(count==gas.size()) return i;
	gasVal = gasVal+gas[j%len]-cost[j%len];
	j++;
	count++;
	}
	else
	{
	i=j;
	break;
	}
	}
	}
	return -1;
	}
	};

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