leetcode--Gas Station
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There are N gas stations along a circular route, where the amount of gas at stationi isgas[i]
.
You have a car with an unlimited gas tank and it costs cost[i]
of gas to travel from stationi 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.
public class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int total = 0;//总油数int cur_sum = 0;//当前油数int start = 0;//出发站for(int i=0;i<gas.length;i++){total += gas[i]-cost[i];//整个旅途的总油数,不断相加就可以了cur_sum += gas[i]-cost[i];//当前油数,每次经过一个站和一段路,就加上这段路消耗的油(也可能是增加的)if(cur_sum<0){//如果当前油数小于0,说明从起始点到现在,是会出现断油情况的,所以我们要换一个起始点cur_sum = 0;//选择新的起始点以后,清空油箱start = (i+1)%gas.length;//选择其实点为当前点的下一个点}}if(total<0) return -1;//如果整个旅途,总油数小于0,说明不存在完成旅途的点/* * 否则返回当前油站位置 * 总的思路是这样的,由于题目要求必须存在这样的一个点,我们在上述遍历过程中,找到最后还会有剩余油的起始点 * 因为如果连这个点开始,都不能走完全程,就不可能存在走完全程的点了(前面已经排除了走不完全程的点) * 所以我们只能返回这个点(虽然这个点只是代表走到n还有剩余的油) */return start; }}
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