134. Gas Station

最新推荐文章于 2025-08-15 16:44:13 发布

转载最新推荐文章于 2025-08-15 16:44:13 发布 · 58 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/yrbbest/p/4438780.html

文章标签：

#数据结构与算法 #java

本文深入解析LeetCode上的经典Gas Station问题，通过运用贪婪算法，阐述了如何找到从一个加油站出发，能够完成环形路线的所有加油站的起始点。文章详细解释了算法的时间复杂度和空间复杂度，并提供了Java实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Hide Tags

Greedy

链接： http://leetcode.com/problems/gas-station/

题解：

经典题gas station，贪婪法。设计一个局部当前的gas，一个全局的gas，当局部gas小于0时，局部gas置零，设置结果为当前index的下一个位置，此情况可能出现多次。当全局gas小于0的话说明没办法遍历所有gas站，返回-1。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        if(gas == null || cost == null || gas.length == 0 || cost.length == 0 || gas.length != cost.length)
            return 0;
        int curGas = 0;
        int totalGas = 0;
        int result = 0;
        
        for(int i = 0; i < gas.length; i++){
            curGas += gas[i] - cost[i];
            totalGas += gas[i] - cost[i];
            if(curGas < 0){
                result = i + 1;
                curGas = 0;
            }
        }
        
        if(totalGas < 0)
            return - 1;
        
        return result;
    }
}

Update:

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        if(gas == null || cost == null || gas.length != cost.length || gas.length == 0)
            return -1;
        int len = gas.length;
        int totalGasRequired = 0, curGas = 0, station = 0;
        
        for(int i = 0; i < len; i++) {
            totalGasRequired += gas[i] - cost[i];
            curGas += gas[i] - cost[i];
            if(curGas < 0) {
                curGas = 0;
                station = i + 1;
            }
        }
        
        return totalGasRequired >= 0 ? station : -1;
    }
}

二刷:

Java:

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        if (gas == null || cost == null || gas.length != cost.length) return -1;
        int totalCost = 0, totalGas = 0, curTank = 0, startingIndex = 0;
        
        for (int i = 0; i < gas.length; i++) {
            totalGas += gas[i];
            totalCost += cost[i];
            curTank += (gas[i] - cost[i]);
            if (curTank < 0) {
                curTank = 0;
                startingIndex = i + 1;
            }
        }
        return totalGas >= totalCost ? startingIndex : -1;
    }
}