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  • Leetcode-Gas Station

    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.

    Analysis:

    The amount of gas left in tank is non-decreasing. If we cannot pass a station P (>x) strating from x, it is because we do not accmulate enough gas in the tank, so we start from x+1, x+2,...P-1, we will still not be able to pass P. Instead, we try to start from P+1, so that we have the maximum accumulative gas in the tank when reaching P.


    If we cannot pass P (<x) starting from x, we already try P+1 (<=x), then we have no way to pass P, fail directly.

    Solution:

     1 public class Solution {
     2     public int canCompleteCircuit(int[] gas, int[] cost) {
     3         if (gas.length==0) return -1;
     4         if (gas.length==1)
     5             if (gas[0]-cost[0]>=0) return 0;
     6             else return -1;
     7 
     8         int len = gas.length;
     9         int left = 0;
    10         int index = 0;
    11         while (index<len){
    12             left = 0;
    13             //Check whether the current station is the solution.
    14             boolean valid = true;
    15             int p = index;
    16             while (p!=(index+gas.length-1)%gas.length){
    17                 if (left+gas[p]-cost[p]<0){
    18                     valid = false;
    19                     break;
    20                 } else{
    21                     left += gas[p]-cost[p];
    22                     p = (p+1)%gas.length;
    23                 }
    24             }
    25             if (valid && left+gas[p]-cost[p]>=0) return index;
    26             else if ((p+1)%gas.length<=index) return -1;
    27             else  index = p+1;
    28         }
    29 
    30         return -1;
    31     }
    32 }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/4179301.html
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