[LeetCode] Gas Station 贪心

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.

Hide Tags

 Greedy

 

---

　　一道贪心算法的题目。

 

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int n = gas.size();
        if(n==0||n!=cost.size())  return -1;
        if(n==1)    return gas[0]>=cost[0]?0:-1;
        int stardIdx =0,endIdx = 0;
        int leave = 0;
        do{
            if(leave+gas[endIdx]>=cost[endIdx]){
                leave = leave+gas[endIdx]-cost[endIdx];
                endIdx++;
                if(endIdx==n)   endIdx = 0;
                continue;
            }
            stardIdx--;
            if(stardIdx==-1)    stardIdx=n-1;
            leave = leave + gas[stardIdx] - cost[stardIdx];
        }while(stardIdx!=endIdx);
        if(leave >=0)   return stardIdx;
        return -1;
    }

};

int main()
{
    vector<int > gas{4};
    vector<int > cost{5};
    Solution sol;
    cout<<sol.canCompleteCircuit(gas,cost)<<endl;
//    for(int i=0;i<gas.size();i++){
//        cout<<gas[i]<<endl;
//    }
    return 0;
}