There are N gas stations along a circular route, where the amount of gas at station i is gas[i]
.javascript
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.java
Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.ui
Note:.net
Example 1:code
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:blog
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.
有n个汽油站,第i个汽油站存有油gas[i],从第i个汽油站前往下一个汽油站的油耗为cost[i],求一辆车可否从某一个汽油站开始(汽车初始油量为0)走遍所有n个汽油站。ip
参见 [leetcode] 134. Gas Station 解题报告。ci
class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int start = 0, left = 0, debt = 0; for (int i = 0; i < gas.length; i++) { left += gas[i] - cost[i]; if (left < 0) { debt += left; start = i + 1; left = 0; } } return left + debt >= 0 ? start : -1; } }
/** * @param {number[]} gas * @param {number[]} cost * @return {number} */ var canCompleteCircuit = function(gas, cost) { let left = [] let remain = 0, owe = 0, start = 0 gas.forEach((item, i) => left.push(item - cost[i])) for (let i = 0; i < gas.length;i++) { remain += left[i] if (remain < 0) { start = i + 1 owe += remain remain = 0 } } if (remain + owe >= 0) { return start } return -1 }