Gas Station

描述

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.

分析

首先想到的是O(N2)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. // Gas Station
  2. // 时间复杂度O(n)，空间复杂度O(1)
  3. public class Solution {
  4.     public int canCompleteCircuit(int[] gas, int[] cost) {
  5.         int total = 0;
  6.         int j = -1;
  7.         for (int i = 0, sum = 0; i < gas.length; ++i) {
  8.             sum += gas[i] - cost[i];
  9.             total += gas[i] - cost[i];
  10.             if (sum < 0) {
  11.                 j = i;
  12.                 sum = 0;
  13.             }
  14.         }
  15.         return total >= 0 ? j + 1 : -1;
  16.     }
  17. };

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/linear-list/array/gas-station.html