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134.gas-station.cpp
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63 lines (61 loc) · 2.3 KB
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// Tag: Array, Greedy
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// 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
//
//
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int n = gas.size();
int total_surplus = 0;
int surplus = 0;
int index = 0;
for (int i = 0; i < n; i++) {
int left = gas[i] - cost[i];
total_surplus += left;
surplus += left;
if (surplus < 0) {
index = i + 1;
surplus = 0;
}
}
return total_surplus >= 0 ? index: -1;
}
};