[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.

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Greedy

 

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　　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==||n!=cost.size())  return -;
        if(n==)    return gas[]>=cost[]?:-;
        int stardIdx =,endIdx = ;
        int leave = ;
        do{
            if(leave+gas[endIdx]>=cost[endIdx]){
                leave = leave+gas[endIdx]-cost[endIdx];
                endIdx++;
                if(endIdx==n)   endIdx = ;
                continue;
            }
            stardIdx--;
            if(stardIdx==-)    stardIdx=n-;
            leave = leave + gas[stardIdx] - cost[stardIdx];
        }while(stardIdx!=endIdx);
        if(leave >=)   return stardIdx;
        return -;
    }

};

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