标签：amount bsp class log int targe 不能 journey start
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.
下边的超时了：
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { if (gas.size() != cost.size() || gas.size() == 0)return -1; int tank = 0; int num = gas.size(); for (int i = 0; i < num;i++) { tank = 0; for (int j = i; j < num + i;j++) { tank += gas[j%num]; tank -= cost[j%num]; if (tank < 0)//不能到下一个station 重新开始 { j = num+i; //退出内层循环 } } if (tank>=0) { return i; } } return -1; }
先放这儿，回头细聊：
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int i, j, n = gas.size(); /* * If start from i, stop before station x -> no station k from i + 1 to x - 1 can reach x. * Bcoz if so, i can reach k and k can reach x, then i reaches x. Contradiction. * Thus i can jump directly to x instead of i + 1, bringing complexity from O(n^2) to O(n). */ // start from station i for (i = 0; i < n; i += j) { int gas_left = 0; // forward j stations for (j = 1; j <= n; j++) { int k = (i + j - 1) % n; gas_left += gas[k] - cost[k]; if (gas_left < 0) break; } if (j > n) return i; } return -1; }
参考：
https://discuss.leetcode.com/topic/8860/fully-commented-o-n-c-solution-enabled-by-a-single-observation-of-mine
标签：amount bsp class log int targe 不能 journey start
原文地址：http://www.cnblogs.com/hellowooorld/p/6481637.html
