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https://leetcode.com/problems/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.
class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int n = gas.size(); if(n == 0) return -1; if(n == 1) { if(gas[0] - cost[0] >= 0) return 0; return -1; } vector<int> dif; dif.clear(); for(int i=0;i<n;++i) dif.push_back(gas[i] - cost[i]); for(int i=0;i<n-1;++i) dif.push_back(gas[i] - cost[i]); vector<int> sum(2*n-1, 0); vector<int> dp(2*n-1, 0); sum[0] = dif[0]; dp[0] = 1; for(int i=1;i<2*n-1;++i) { if(sum[i-1] < 0) { sum[i] = dif[i]; dp[i] = 1; } else { sum[i] = sum[i-1] + dif[i]; dp[i] = dp[i-1] + 1; } if(dp[i] == n && sum[i] >= 0) return i-n+1; } return -1; } };
leetcode@ [134] Gas station (Dynamic Programming)
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原文地址：http://www.cnblogs.com/fu11211129/p/5103305.html
