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  • Gas Station [LeetCode]

    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.

    Solution:

    The brute force solution costs O(n^2), I find a O(n) and one round solution. If gas[i] >= cost[i], then I mark it, and calcuate the gas in tank along next node, if it is less than zero, then restart search.

     1     int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
     2         int curr_gas = 0;
     3         int idx = -1;
     4         int tmp_gas = 0;
     5         for(int i = 0; i < gas.size(); i ++) {
     6             curr_gas += gas[i] - cost[i];
     7             if( idx != -1)
     8                 tmp_gas += gas[i] - cost[i];
     9             
    10             if(tmp_gas < 0) {
    11                 idx = -1;
    12                 tmp_gas = 0;   
    13             }
    14             
    15             if(gas[i] - cost[i] >= 0 && idx == -1) {
    16                 idx = i;
    17                 tmp_gas += gas[i] - cost[i];
    18             }
    19         }
    20         
    21         if(curr_gas >= 0)
    22             return idx;
    23         else 
    24             return -1;
    25     }
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  • 原文地址:https://www.cnblogs.com/guyufei/p/3425175.html
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