LeetCode-----------gas-station

There are N gas stations along a circular route, where the amount of gas at station i isgas[i].

You have a car with an unlimited gas tank and it costscost[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.

提交的代码

语言：C++ 运行时间： <1 ms 占用内存：8568K 状态：答案正确

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class Solution { public:     int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {         int N = cost.size();         int i, j;         int alc, alg;//当前总cost和当前总gas                   for (i=0; i<N; i++){//外循环i站             if (cost[i] > gas[i])                 continue;             if (i == N-1)                 j = 0;             else                 j = i+1;             alc = cost[i];             alg = gas[i];                           while (j!=i){//内循环判断第i站能否做起点。由当前总cost<当前总gas来保证                 alc +=cost[j];                 alg +=gas[j];                                   if (alg >= alc){//能坚持到下一站                                           if (j == N-1){                         j = 0;                                         }else{                         j++;                     }                                       }                 else break;             }             if (j==i)                 return i;         }         return -1;     } };

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借鉴方案：

1 2	从start出发， 如果油量足够， 可以一直向后走 end++； 油量不够的时候，start向后退  最终 start == end的时候，如果有解一定是当前 start所在位置

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class Solution { public:  int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {               int start = gas.size() - 1;         int end = 0;         int sum = gas[start] - cost[start];         while(start > end){             if(sum >= 0){                 sum += gas[end] - cost[end];                 ++end;             }else{                 --start;                 sum += gas[start] - cost[start];             }         }         return sum >=0 ? start : -1;                         } };