Gas Station 187

Question

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.

Notice

The solution is guaranteed to be unique.

Example

Given 4 gas stations with gas[i]=[1,1,3,1], and the cost[i]=[2,2,1,1]. The starting gas station's index is 2.

Challenge

O(n) time and O(1) extra space

Solution

这道题只要看gas和是否小于cost和，若小于则一定没有解，反之则一定有解，此时只要找到开头（即当前gas的值加上之前剩下的gas的值大于等于当前cost的值，即能够到达下一个station），看其到数组最后一个station中间的每个点是否都满足条件（当前gas的值加上之前剩下的gas的值大于等于当前cost的值），若满足则返回此开头，否则从不满足的点的下一个点开始重新寻找开头。该问题属于贪心算法。

代码如下：

public class Solution {
    /**
     * @param gas: an array of integers
     * @param cost: an array of integers
     * @return: an integer
     */
    //若gas和小于cost和则一定没有解，反之则一定有解
    public int canCompleteCircuit(int[] gas, int[] cost) {
        // write your code here
        if(gas == null || gas.length == 0 || cost == null || cost.length == 0 || gas.length != cost.length){
            return -1;
        }

        int n = gas.length;
        int sumGas = 0;
        int sumCost = 0;
        for(int i = 0; i < n; i++){
            sumGas += gas[i];
            sumCost += cost[i];
        }
        if(sumGas < sumCost){
            return -1;
        }

        int start = 0;
        int diff = 0;
        for(int i = 0; i < n; i++){
            if(diff + gas[i] < cost[i]){
                start = i + 1;
                diff = 0;
            }else{
                diff += gas[i] - cost[i];
            }
        }

        return start;
    }
}