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.

$O(N)$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。

// 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。