Leetcode Week7 Gas Station

Question

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.

Answer

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        vector<int> fuel(gas.size());
        for (int i = 0; i < gas.size(); i++) {
            fuel[i] = gas[i]-cost[i];
        }
        int left, right, maxfuel;
        maxfuel = maxSubarraySumCircular(fuel, left, right);
        if (left <= right) {
            for (int i = right+1; i < gas.size(); i++) {
                maxfuel += fuel[i];
                if (maxfuel < 0) return -1;
            }
             for (int i = 0; i < left; i++) {
                maxfuel += fuel[i];
                if (maxfuel < 0) return -1;
            }
            return left;
        } else {
            for (int i = right+1; i < left; i++) {
                maxfuel += fuel[i];
                if (maxfuel < 0) return -1;
            }
            return left;
        }
    }
         int maxSubarraySumCircular(vector<int>& A, int &left, int &right) {
            int total = 0, maxSum = -30000, curMax = 0, minSum = 30000, curMin = 0, Start = 0, Start2 = 0, End = 0, maxStart = 0, maxEnd = 0, 
             minStart = 0, minEnd = 0;
            for (int i = 0; i < A.size(); i++) {
                int a = A[i];
                if (curMax < 0) {
                    Start = i;
                }
                curMax = max(curMax + a, a);
                if (maxSum < curMax) {
                    maxStart = Start;
                    maxEnd = i;
                }
                maxSum = max(maxSum, curMax);
                if (curMin > 0) {
                    Start2 = i;
                }
                curMin = min(curMin + a, a);
                if (minSum > curMin) {
                    minEnd = i;
                    minStart = Start2;
                }
                minSum = min(minSum, curMin);
                total += a;
            }
             if (maxSum >= total-minSum) {
                 left = maxStart;
                 right = maxEnd;
             } else {
                 left = minEnd+1;
                 right = minStart-1;
             }
            return maxSum > 0 ? max(maxSum, total - minSum) : maxSum;
    }

};

　　首先计算gas和cost数组的差，得到一个有正有负的数列，然后求连续的和最大的子数列（同时也要保存最大子数列的起始位置 = start），可以转化为Find Minimum in Rotated Sorted Array II的问题。