首页 > ACM题库 > LeetCode > LeetCode-Gas Station[数组]
2014
11-19

LeetCode-Gas Station[数组]

Gas Station

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.

标签: Greedy
分析

首先想到的是$O(N^2)$的解法,对每个点进行模拟。

$O(N)$的解法是,设置两个变量,{sum}判断当前的指针的有效性;{total}则判断整个数组是否有解,有就返回通过{sum}得到的下标,没有则返回-1。

代码1

// LeetCode, Gas Station
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int total = 0;
        int j = -1;
        for (int i = 0, sum = 0; i < gas.size(); ++i) {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            if (sum < 0) {
                j = i;
                sum = 0;
            }
        }
        return total >= 0 ? j + 1 : -1;
    }
};

Java代码:

class Solution:
    # @param gas, a list of integers
    # @param cost, a list of integers
    # @return an integer
    def canCompleteCircuit(self, gas, cost):
        n = len(gas)
        pre = 0
        curr_sum = 0
        total = 0
        for i in range(n):
            total += gas[i] - cost[i]
            curr_sum += gas[i] - cost[i]
            if curr_sum < 0:
                curr_sum = 0
                pre = i+1
        if total < 0:
            return -1
        else:
            return pre

  1. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。