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  • LeetCode 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.

    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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/birdhack/p/3956684.html
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