Question

134. Gas Station

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Description

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return _the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return_ -1. If there exists a solution, it is guaranteed to be unique

 

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.

 

Constraints:

• n == gas.length == cost.length
• 1 <= n <= 105
• 0 <= gas[i], cost[i] <= 104

Solutions

Solution 1

class Solution:
  def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
    n = len(gas)
    i = j = n - 1
    cnt = s = 0
    while cnt < n:
      s += gas[j] - cost[j]
      cnt += 1
      j = (j + 1) % n
      while s < 0 and cnt < n:
        i -= 1
        s += gas[i] - cost[i]
        cnt += 1
    return -1 if s < 0 else i

class Solution {
  public int canCompleteCircuit(int[] gas, int[] cost) {
    int n = gas.length;
    int i = n - 1, j = n - 1;
    int cnt = 0, s = 0;
    while (cnt < n) {
      s += gas[j] - cost[j];
      ++cnt;
      j = (j + 1) % n;
      while (s < 0 && cnt < n) {
        --i;
        s += gas[i] - cost[i];
        ++cnt;
      }
    }
    return s < 0 ? -1 : i;
  }
}

class Solution {
public:
  int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
    int n = gas.size();
    int i = n - 1, j = n - 1;
    int cnt = 0, s = 0;
    while (cnt < n) {
      s += gas[j] - cost[j];
      ++cnt;
      j = (j + 1) % n;
      while (s < 0 && cnt < n) {
        --i;
        s += gas[i] - cost[i];
        ++cnt;
      }
    }
    return s < 0 ? -1 : i;
  }
};

func canCompleteCircuit(gas []int, cost []int) int {
  n := len(gas)
  i, j := n-1, n-1
  cnt, s := 0, 0
  for cnt < n {
    s += gas[j] - cost[j]
    cnt++
    j = (j + 1) % n
    for s < 0 && cnt < n {
      i--
      s += gas[i] - cost[i]
      cnt++
    }
  }
  if s < 0 {
    return -1
  }
  return i
}

function canCompleteCircuit(gas: number[], cost: number[]): number {
  const n = gas.length;
  let i = n - 1;
  let j = n - 1;
  let s = 0;
  let cnt = 0;
  while (cnt < n) {
    s += gas[j] - cost[j];
    ++cnt;
    j = (j + 1) % n;
    while (s < 0 && cnt < n) {
      --i;
      s += gas[i] - cost[i];
      ++cnt;
    }
  }
  return s < 0 ? -1 : i;
}

public class Solution {
  public int CanCompleteCircuit(int[] gas, int[] cost) {
    int n = gas.Length;
    int i = n - 1, j = n - 1;
    int s = 0, cnt = 0;
    while (cnt < n) {
      s += gas[j] - cost[j];
      ++cnt;
      j = (j + 1) % n;
      while (s < 0 && cnt < n) {
        --i;
        s += gas[i] - cost[i];
        ++cnt;
      }
    }
    return s < 0 ? -1 : i;
  }
}