Gas Station, Leetcode 解题笔记

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.

感觉更像是数学题,佩服那些能想到下面这种解法的人:
http://fisherlei.blogspot.com/2013/11/leetcode-gas-station-solution.html
蛮精巧的一道题。最直白的解法就是从每一个点开始,遍历整个环,然后找出最后剩余油量最大的点。这个是O(n^2)的。但是这题明显不会无聊到让做题人写个两层循环这么简单。

仔细想一下,其实和以前求最大连续子数组和的题很像。

在任何一个节点,其实我们只关心油的损耗,定义:

diff[i] = gas[i] – cost[i] 然后对于这个区间的diff加和,定义

sum[i,j] = ∑diff[k]

如果sum[i,j]小于0,那么这个起点肯定不会在[i,j]这个区间里,跟第一个问题的原理一样。举个例子,假设i是[0,n]的解,那么我们知道 任意sum[k,i-1]肯定是小于0的,否则解就应该是k。同理,sum[i,n]一定是大于0的,否则,解就不应该是i,而是i和n之间的某个点。所以第二题的答案,其实就是在0到n之间,找到第一个连续子序列(这个子序列的结尾必然是n)大于0的。

至此,两个问题都可以在一个循环中解决。

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int[] diff = new int[gas.length];
        int total_diff = 0;
        int sum = 0;
        int ret = 0;
        for(int i = 0; i < gas.length; i++){
            diff[i] = gas[i] - cost[i];
            total_diff += diff[i];
            sum += diff[i];
            if(total_diff < 0){
                total_diff = 0;
                ret = i+1;
            }
        }
        if(sum < 0) return -1;
        else return ret;
    }
}
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