鸿 网 互 联 www.68idc.cn

当前位置 : 服务器租用 > 编程语言开发 > erlang > >

【Leetcode】Gas Station

来源:互联网 作者:佚名 时间:2016-05-30 10:49
题目链接:https://leetcode.com/problems/gas-station/ 题目: There areNgas stations along a circular route, where the amount of gas at stationiis gas[i] . You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from

题目链接:https://leetcode.com/problems/gas-station/
题目:

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.

思路:

看了tag发现是贪心,想起老师上课说贪心实现起来是最简单的,最难的是证明贪心策略的正确性。这里主要为了AC,多尝试几次可能的策略懒得证明了(宝宝TM不会证明啊!!)。很自然想到的策略有:1、从gas最多的station开始 2、从cost最少的station开始 3、从剩余(gas-cost)最少的开始。

尝试了几次发现第二种是对的,但要注意如果有多个站点cost相同,要判断这些站点是否能完成环圈旅行。

算法:

public int canCompleteCircuit(int[] gas, int[] cost) {  
        int index = 0;  
          
        for (int i = 1; i < cost.length; i++) {  
            if (cost[i]<cost[index]) {  
                index = i;  
            }  
        }  
          
        for(int i=0;i<cost.length;i++){  
            if(cost[i]==cost[index]){   
                if(isValid(cost, gas, i)!=-1){  
                    return i;  
                }  
            }  
        }  
        return -1;  
    }  
      
    public int isValid(int[]cost,int[] gas,int index){  
        int tankGas = 0;  
        for(int i=index;i<gas.length;i++){  
            tankGas+=gas[i]-cost[i];  
            if(tankGas<0){  
                return -1;  
            }  
        }  
        for(int i=0;i<index;i++){  
            tankGas+=gas[i]-cost[i];  
            if(tankGas<0){  
                return -1;  
            }  
        }  
        return index;  
    }  


网友评论
<