题解 | #加油站#

import java.util.*;

/**
 * NC235 加油站
 * @author d3y1
 */
public class Solution {
    private int n;

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param gas int整型一维数组
     * @param cost int整型一维数组
     * @return int整型
     */
    public int gasStation (int[] gas, int[] cost) {
        return solution1(gas, cost);
        // return solution2(gas, cost);
    }

    /**
     * 贪心
     * 
     *                 start
     *                   |
     *      i:  0  1  2  3  4
     *    gas:  4  1  2  5  3
     *   cost:  1  3  4  2  5
     *   diff:  3 -2 -2  3 -2
     * remain:  3  1 -1  3  1
     *    sum:  3  1 -1  2  0
     *    
     * R-remain D-diff start=0
     * R00 = D0 = 3
     * R01 = D0+D1 = 3+(-2) = 1
     * R02 = D0+D1+D2 = 3+(-2)+(-2) = -1, 小于0 => start = i+1 = 2+1 = 3
     * start可从0直接到3
     * 
     * 推理:
     * 若从1开始(start=1)
     * D0 = 3 >= 0
     * => D0+D1 >= D1 (从0开始尚且不行, 从1开始更是不行)
     * 
     * 若从2开始(start=2)
     * D0+D1 = 1 >= 0
     * D0+D1+D2 = -1 < 0
     * => D2 < 0 (从2开始也是不行)
     * 
     * @param gas
     * @param cost
     * @return
     */
    private int solution1(int[] gas, int[] cost){
        n = gas.length;

        int start = 0;
        int remain = 0;
        int sum = 0;
        int diff;
        for(int i=0; i<n; i++){
            diff = gas[i]-cost[i];
            remain += diff;
            sum += diff;
            if(remain < 0){
                start = (i+1)%n;
                remain = 0;
            }
        }

        return sum>=0?start:-1;
    }

    /**
     * 循环遍历
     * @param gas
     * @param cost
     * @return
     */
    private int solution2(int[] gas, int[] cost){
        n = gas.length;

        for(int i=0; i<n; i++){
            if(gas[i] >= cost[i]){
                if(canTravel(gas, cost, i)){
                    return i;
                }
            }
        }

        return -1;
    }

    /**
     * 能否绕环路行驶一周
     * @param gas
     * @param cost
     * @param start 出发的加油站编号
     * @return
     */
    private boolean canTravel(int[] gas, int[] cost, int start){
        int remain = gas[start]-cost[start];
        int next = (start+1)%n;
        while(next != start){
            remain += (gas[next]-cost[next]);
            if(remain < 0){
                return false;
            }
            next = (next+1)%n;
        }

        return true;
    }
}