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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.
Greedy
#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; }
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原文地址：http://www.cnblogs.com/Azhu/p/4332812.html
