leetcode:Gas Station

最新推荐文章于 2021-02-13 03:33:59 发布

原创最新推荐文章于 2021-02-13 03:33:59 发布 · 418 阅读

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解决环路加油站问题，确定从哪个加油站出发能使车辆绕环路一周。通过两种算法实现：一种直接遍历每个起点检查是否可行；另一种使用优化方法，合并两次遍历的数据减少计算复杂度。

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.

下面的解法大数据时超时

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        
        int currentSum = 0;
        int size = gas.size();
        
        for (int i=0; i<size; i++)
        {
            currentSum = 0;
            int j;
            for (j=i; j<i+size; j++)
            {
                if (currentSum+gas[(j+size)%size] < cost[(j+size)%size])
                    break;
                    
                currentSum = currentSum+gas[(j+size)%size]-cost[(j+size)%size];
            }
            if (j == i+size)
                return j;
        }
        
        return -1;
    }
};

另外一个Accept的解法

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        
        int size = gas.size();
        vector<int> merged(2*size);
        
        for (int i=0; i<2*size; i++)
        {
            merged[i] = gas[i%size]-cost[i%size];
        }
        
        int curSum  = 0;
        int startPos = -1;
        for(int i=0; i<2*size && startPos<size; i++)
        {
            if (startPos == -1)
            {
                startPos = i;            
            }
            
            curSum += merged[i];

            if (curSum < 0)
            {
                curSum = 0;
                startPos = -1;
                continue;
            }
            
            if (i-startPos+1 == size)
            {
                break;
            }
        }
        
        if (startPos != -1 && startPos < size)
        {
            return startPos;
        }
        
        return -1;
    }
};