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.
把路径上所有值加起来，找到第一个，从该点到整个路径的和大于0的下标；如果不存在这样的点， 就输出-1；
int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize) { int index = 0; int suc_sum = 0; int add_sum = 0; // go over the gas[] and cost[] for(int i = 0; i < gasSize; i++){ int j = gas[i] - cost[i]; // record add sum and the first Non-negative index and its successive sum add_sum += j; suc_sum += j; if(suc_sum < 0){ index = i + 1; suc_sum = 0; } } // if sum is Non-negative return the first Non-negatice number if(add_sum >= 0){ return index; }else{ // else return -1 return -1; } }
原文：http://www.cnblogs.com/dylqt/p/5077615.html