leetcode 134. Gas Station ----- java

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, otherwise return -1.

Note:
The solution is guaranteed to be unique.

1、暴力的做法

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int len = gas.length;
        int sum = 0;
        int[] dp = new int[len];
        for( int i = 0;i<len;i++){
            dp[i] = gas[i]-cost[i];
            sum+=dp[i];
        }
        int num = 0;
        if( sum < 0)
            return -1;
        for( int i = 0;i<len;i++){
            num = 0;
            for( int j = i;j<len;j++){
                num+=dp[j];
                if( num < 0){
                    break;
                }
            }
            if( num >= 0 )
                return i;
        }

        return -1;

    }
}

2、仔细看题目，发现答案其实是唯一解，那么就只需要从后向前遍历，找到最大子串和的起始位置i,（结束位置是最后一位）

在总消耗>0的前提下，返回起始位置i。

否则返回-1。

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {

        int len = gas.length;
        int sum = 0;
        int max = 0;
        int start = 0;
        for( int i = len-1;i>=0;i--){
            sum+=(gas[i]-cost[i]);
            if( sum > max ){
                max = sum;
                start = i;
            }
        }
        if( sum < 0)
            return -1;
        return start;

    }
}