LeetCode 134. Gas Station 题解_there are n gas stations along a circular route-优快云博客

博客介绍了LeetCode 134题的解法，重点讨论了遍历数组gas和cost来判断汽车是否能完成环形路线的思路。通过计算每个起点到终点的净油量，如果总和大于0则表示可以从该站出发完成循环，这种方法的时间复杂度为O(n^2)。

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【题目】

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.

【题解】

最容易想到的解法是遍历数组gas和数组cost每个索引 i，即以索引 i 为起点，将后面的数(gas[j] - cost[j])相加(循环回到 i )，如果最后得到的结果nTotal大于0，证明车从索引 i 开始可以循环一圈，时间复杂度为O(n^2)。

以上是我很愚蠢的想法，在参考了网上大神的思路后，发觉可以将复杂度降为O(n)，很惭愧，真的很简单，在这里我就不重复这位大神的思路了，需要的话可以参考点击打开链接

【代码】

int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int gasSize = gas.size();
        int costSize = cost.size();
        if (gasSize == 0 || costSize == 0 || gasSize != costSize)
            return -1;
            
        int nTotal = 0;
        int nBegin = 0;
        int nSum = 0;
        int nDiff = 0;
        for (int i = 0; i < gasSize; i++) {
            nDiff = gas[i] - cost[i];
            nTotal += nDiff;
            if (nSum < 0) {
                nSum = nDiff;
                nBegin = i;
            }
            else
                nSum += nDiff;
        }
        
        if (nTotal < 0)
            return -1;
        
        return nBegin;
    }