【Leetcode】Gas Station

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;
}