leetcode@ [134] gas station (Dynamic programming)

Source: Internet
Author: User

https://leetcode.com/problems/gas-station/

There was N gas stations along a circular route, where the amount of gas at Station I was `gas[i]` .

You had a car with an unlimited gas tank and it costs of gas to travel from station `cost[i]` I to its next station (i+ 1). You begin the journey with a 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.

`classSolution { Public:    intCancompletecircuit (vector<int>& Gas, vector<int>&Cost ) {        intn =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(intI=0; i<n;++i) Dif.push_back (Gas[i]-Cost[i]);  for(intI=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(intI=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)returni-n+1; }                return-1; }};`
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[email protected] [134] Gas station (Dynamic programming)

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