134-Gas Station

100-Same Tree | Links:

题意

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 in the clockwise direction, otherwise return -1. Note:

• If there exists a solution, it is guaranteed to be unique.
• Both input arrays are non-empty and have the same length.
• Each element in the input arrays is a non-negative integer.

Example 1:
Input: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

Output: 3

Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

Example 2:
Input: 
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

思路

• 贪心算法
• 数学定理：如果一个数组总和非负，那么一定可以找到一个起始位置，从此处开始遍历一圈，累加和一直都是非负的
• 找连续和>0：把全部的油耗情况计算出来看看是否大于等于0即可

class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        if not gas or not cost or len(gas)!= len(cost): return -1
        start = 0 # 记录当前起点
        remain = 0 # 当前tank剩余
        debt = 0 # 从0开始走的欠债
        
        for i in range(len(gas)):
            remain += gas[i] - cost[i] # 这一站加油 - 到下一站花费 -> 没站积累
            if remain < 0:
                start = i + 1 # 从下一站开始继续验证
                debt += remain # 把前面欠债记录下来，避免重复计算
                remain = 0 # 重置邮箱
        
        return start if remain + debt >= 0 else -1

分析:

• Time: O(n)
• Space: O(1)