leetCode(54):Gas Station

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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.

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int length=gas.size();
    	vector<int> tmpGas(length,0);
    	
    	for(int i=0;i<length;++i)
    	{
    		tmpGas[i]=gas[i]-cost[i];//两者之差组成的新数组，代表补充或消耗
    	}
    	
    	int startIndex=0;//初始起点
    	int sum=tmpGas[startIndex];
    	
    	int i=0;	
    	while(true)
    	{
    		i++;	
    		if(i>=length)
    		{
    			i=i-length;
    		}
    		
    		if(sum<0)
    		{
    			startIndex=i;//以下一个节点作为起点
    			if(startIndex==0)
    				return -1;//又回到初始起点，开始出现循环
    			sum=tmpGas[startIndex];
    		}
    		else
    		{
    			sum+=tmpGas[i];

<span style="white-space:pre">				</span>
    			//已经到达终点，判断是否能够顺利开完全程
    			if(startIndex==0 && i==length-1)
    			{//如果起点是0，当前结点又是最后一个节点
    				if(sum<0)
    					return -1;
    				else
    					return startIndex;
    			}    
    			if(i==startIndex-1)
    			{//如果当前结点是循环的最后一个结点
    				if(sum>=0)
    					return startIndex;
    				else
    					return -1;					
    			}
    		}
    	}
    }
};

leetCode(54):Gas Station

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原文地址：http://blog.csdn.net/walker19900515/article/details/47312155