标签：style blog color os io ar strong for art
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.
题目意思比较简单，一圈加油站，每个加油站的汽油是gas[]，到下一个加油站的消耗是cost[]。给出gas[] cost[]，算出从哪个加油站出发可以跑完一圈。
解法就是遍历每一个加油站，每次循环内分成两段，设i开始，到末尾结束，然后从0开始到i结束。依次累加，判断剩余油sum+ gas[]-cost[],若中间有为负数的，就失败。
public class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int sum=0; boolean nextloop = false; for (int i = 0; i < cost.length; i++) { for (int j = i; j < cost.length; j++) { sum=sum+gas[j]-cost[j]; if (sum<0) { nextloop=false; break; }else { nextloop=true; } } if (nextloop) { for (int j = 0; j < i; j++) { sum=sum+gas[j]-cost[j]; if (sum<0) { nextloop=false; break; } } } if (sum>=0) { return i; } sum=0; } return -1; } }
标签：style blog color os io ar strong for art
原文地址：http://www.cnblogs.com/birdhack/p/3956684.html
